Solution

Let $x^{3}=t$

$\therefore 3 x^{2} d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{3 x^{2}}{x^{6}+1} d x & =\int \dfrac{d t}{t^{2}+1} \\ & =\tan ^{-1} t+C \\ & =\tan ^{-1}(x^{3})+C \end{aligned} $

Solution

Let $2 x=t$

$\therefore 2 d x=d t$

$\Rightarrow \int \dfrac{1}{\sqrt{1+4 x^{2}}} d x=\dfrac{1}{2} \int \dfrac{d t}{\sqrt{1+t^{2}}}$

$ \begin{matrix} =\dfrac{1}{2}[\log |t+\sqrt{t^{2}+1}|]+C & {[\because \int \dfrac{1}{\sqrt{x^{2}+a^{2}}} d t=\log |x+\sqrt{x^{2}+a^{2}}|]} \\ =\dfrac{1}{2} \log |2 x+\sqrt{4 x^{2}+1}|+C & \end{matrix} $

Solution

Let $2-x=t$

$\Rightarrow-d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{1}{\sqrt{(2-x)^{2}+1}} d x & =-\int \dfrac{1}{\sqrt{t^{2}+1}} d t \\ & =-\log |t+\sqrt{t^{2}+1}|+C \quad[\because\int \dfrac{1}{\sqrt{x^{2}+a^{2}}} d t=\log |x+\sqrt{x^{2}+a^{2}}|+ C] \\ & =-\log |2-x+\sqrt{(2-x)^{2}+1}|+C \\ & =\log |\dfrac{1}{(2-x)+\sqrt{x^{2}-4 x+5}}|+C \end{aligned} $

Solution

Let $5 x=t$

$\therefore 5 d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{1}{\sqrt{9-25 x^{2}}} d x & =\dfrac{1}{5} \int \dfrac{1}{9-t^{2}} d t \\ & =\dfrac{1}{5} \int \dfrac{1}{\sqrt{3^{2}-t^{2}}} d t \\ & =\dfrac{1}{5} \sin ^{-1}(\dfrac{t}{3})+C \\ & =\dfrac{1}{5} \sin ^{-1}(\dfrac{5 x}{3})+C \end{aligned} $

Solution

Let $\sqrt{2} x^{2}=t$

$\therefore 2 \sqrt{2} x d x=d t$

$\Rightarrow \int \dfrac{3 x}{1+2 x^{4}} d x=\dfrac{3}{2 \sqrt{2}} \int \dfrac{d t}{1+t^{2}}$

$ =\dfrac{3}{2 \sqrt{2}}[\tan ^{-1} t]+C $

$=\dfrac{3}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} x^{2})+C$

Solution

Let $x^{3}=t$

$\therefore 3 x^{2} d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{x^{2}}{1-x^{6}} d x & =\dfrac{1}{3} \int \dfrac{d t}{1-t^{2}} \\ & =\dfrac{1}{3}[\dfrac{1}{2} \log |\dfrac{1+t}{1-t}|]+C \\ & =\dfrac{1}{6} \log |\dfrac{1+x^{3}}{1-x^{3}}|+C \end{aligned} $

Solution

$ \begin{aligned} \int \dfrac{x-1}{\sqrt{x^{2}-1}} d x=\int \dfrac{x}{\sqrt{x^{2}-1}} d x-\int \dfrac{1}{\sqrt{x^{2}-1}} d x \qquad…(1) \end{aligned} $

For $\int \dfrac{x}{\sqrt{x^{2}-1}} d x$, let $x^{2}-1=t \Rightarrow 2 x d x=d t$

$\therefore \int \dfrac{x}{\sqrt{x^{2}-1}} d x=\dfrac{1}{2} \int \dfrac{d t}{\sqrt{t}}$

$=\dfrac{1}{2} \int t^{-\dfrac{1}{2}} d t$

$=\dfrac{1}{2}[2 t^{\dfrac{1}{2}}]$

$=\sqrt{t}$

$=\sqrt{x^{2}-1}$

From (1), we obtain

$ \begin{aligned} \int \dfrac{x-1}{\sqrt{x^{2}-1}} d x & =\int \dfrac{x}{\sqrt{x^{2}-1}} d x-\int \dfrac{1}{\sqrt{x^{2}-1}} d x \quad \quad[\because \int \dfrac{1}{\sqrt{x^{2}-a^{2}}} d t=\log |x+\sqrt{x^{2}-a^{2}}|] \\ & =\sqrt{x^{2}-1}-\log |x+\sqrt{x^{2}-1}|+C \end{aligned} $

Solution

Let $x^{3}=t$

$\Rightarrow 3 x^{2} d x=d t$

$ \begin{aligned} \therefore \int \dfrac{x^{2}}{\sqrt{x^{6}+a^{6}}} d x & =\dfrac{1}{3} \int \dfrac{d t}{\sqrt{t^{2}+(a^{3})^{2}}} \\ & =\dfrac{1}{3} \log |t+\sqrt{t^{2}+a^{6}}|+C \\ & =\dfrac{1}{3} \log |x^{3}+\sqrt{x^{6}+a^{6}}|+C \end{aligned} $

Solution

Let $\tan x=t$

$\therefore \sec ^{2} x d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}} d x & =\int \dfrac{d t}{\sqrt{t^{2}+2^{2}}} \\ & =\log |t+\sqrt{t^{2}+4}|+C \\ & =\log |\tan x+\sqrt{\tan ^{2} x+4}|+C \end{aligned} $

Solution

$ \int \dfrac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \dfrac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x $

Let $x+1=t$

$\therefore d x=d t$

$\Rightarrow \int \dfrac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \dfrac{1}{\sqrt{t^{2}+1}} d t$

$ \begin{aligned} & =\log |t+\sqrt{t^{2}+1}|+C \\ & =\log |(x+1)+\sqrt{(x+1)^{2}+1}|+C \\ & =\log |(x+1)+\sqrt{x^{2}+2 x+2}|+C \end{aligned} $

Solution

$ \int \dfrac{1}{9 x^{2}+6 x+5} d x=\int \dfrac{1}{(3 x+1)^{2}+(2)^{2}} d x $

Let $(3 x+1)=t$

$\therefore 3 d x=d t$

$\Rightarrow \int \dfrac{1}{(3 x+1)^{2}+(2)^{2}} d x=\dfrac{1}{3} \int \dfrac{1}{t^{2}+2^{2}} d t$

$ \begin{aligned} & =\dfrac{1}{3}[\dfrac{1}{2} \tan ^{-1}(\dfrac{t}{2})]+C \\ & =\dfrac{1}{6} \tan ^{-1}(\dfrac{3 x+1}{2})+C \end{aligned} $

To integrate the function $\dfrac{1}{\sqrt{7-6x-x^2}}$, we can start by completing the square in the denominator.

First, consider the quadratic expression $7 - 6x - x^2$. We can rewrite it as follows:

$7 - 6x - x^2 = -((x + 3)^2 - 16) = 16 - (x + 3)^2$

Thus, the integral becomes:

$ \int \dfrac{1}{\sqrt{16 - (x + 3)^2}} , dx $

This integral is of the form $\int \dfrac{1}{\sqrt{a^2 - u^2}} , du$, which is a standard integral that evaluates to $\sin^{-1}\left(\dfrac{u}{a}\right) + C$.

Here, $a = 4$ and $u = x + 3$. Therefore, the integral is:

$ \int \dfrac{1}{\sqrt{16 - (x + 3)^2}} , dx = \sin^{-1} \left(\dfrac{x + 3}{4}\right) + C $

where $C$ is the constant of integration.

So, the final answer is:

$ \int \dfrac{1}{\sqrt{7-6x-x^2}} , dx = \sin^{-1}\left(\dfrac{x + 3}{4}\right) + C $

Solution

$(x-1)(x-2)$ can be written as $x^{2}-3 x+2$.

Therefore,

$x^{2}-3 x+2$

$=x^{2}-3 x+\dfrac{9}{4}-\dfrac{9}{4}+2$

$=(x-\dfrac{3}{2})^{2}-\dfrac{1}{4}$

$=(x-\dfrac{3}{2})^{2}-(\dfrac{1}{2})^{2}$

$\therefore \int \dfrac{1}{\sqrt{(x-1)(x-2)}} d x=\int \dfrac{1}{\sqrt{(x-\dfrac{3}{2})^{2}-(\dfrac{1}{2})^{2}}} d x$

Let $x-\dfrac{3}{2}=t$

$\therefore d x=d t$

$\Rightarrow \int \dfrac{1}{\sqrt{(x-\dfrac{3}{2})^{2}-(\dfrac{1}{2})^{2}}} d x=\int \dfrac{1}{\sqrt{t^{2}-(\dfrac{1}{2})^{2}}} d t$

$=\log |t+\sqrt{t^{2}-(\dfrac{1}{2})^{2}}|+C$

$=\log |(x-\dfrac{3}{2})+\sqrt{x^{2}-3 x+2}|+C$

Solution

$8+3 x-x^{2}$ can be written as $8-(x^{2}-3 x+\dfrac{9}{4}-\dfrac{9}{4})$.

Therefore,

$8-(x^{2}-3 x+\dfrac{9}{4}-\dfrac{9}{4})$

$=\dfrac{41}{4}-(x-\dfrac{3}{2})^{2}$

$\Rightarrow \int \dfrac{1}{\sqrt{8+3 x-x^{2}}} d x=\int \dfrac{1}{\sqrt{\dfrac{41}{4}-(x-\dfrac{3}{2})^{2}}} d x$

Let $x-\dfrac{3}{2}=t$

$\therefore d x=d t$

$\Rightarrow \int \dfrac{1}{\sqrt{\dfrac{41}{4}-(x-\dfrac{3}{2})^{2}}} d x=\int \dfrac{1}{\sqrt{(\dfrac{\sqrt{41}}{2})^{2}-t^{2}}} d t$

$=\sin ^{-1}(\dfrac{t}{\dfrac{\sqrt{41}}{2}})+C$

$=\sin ^{-1}(\dfrac{x-\dfrac{3}{2}}{\dfrac{\sqrt{41}}{2}})+C$

$=\sin ^{-1}(\dfrac{2 x-3}{\sqrt{41}})+C$

Solution

$\begin{aligned} (x - a)(x - b) &= x^2 - (a + b)x + ab = x^2 - (a + b)x + \dfrac{(a + b)^2}{4} - \dfrac{(a + b)^2}{4} + ab \ &= \left[ x - \dfrac{a + b}{2} \right]^2 - \dfrac{(a - b)^2}{4} \ &\Rightarrow \int \dfrac{1}{\sqrt{(x - a)(x - b)}} , dx = \int \dfrac{1}{\sqrt{\left[ x - \dfrac{a + b}{2} \right]^2 - \dfrac{(a - b)^2}{4}}} , dx \ \text{Let } x - \dfrac{a + b}{2} &= t \Rightarrow dx = dt \ &\Rightarrow \int \dfrac{1}{\sqrt{\left[ x - \dfrac{a + b}{2} \right]^2 - \dfrac{(a - b)^2}{4}}} , dx = \int \dfrac{1}{\sqrt{t^2 - \left( \dfrac{a - b}{2} \right)^2}} , dt \ &= \log \left| t + \sqrt{t^2 - \left( \dfrac{a - b}{2} \right)^2} \right| + C \ &= \log \left| x - \dfrac{a + b}{2} + \sqrt{(x - a)(x - b)} \right| + C \end{aligned}$

Solution

Let $4 x+1=A \dfrac{d}{d x}(2 x^{2}+x-3)+B$

$\Rightarrow 4 x+1=A(4 x+1)+B$

$\Rightarrow 4 x+1=4 A x+A+B$

Equating the coefficients of $x$ and constant term on both sides, we obtain

$4 A=4 \Rightarrow A=1$

$A+B=1 \Rightarrow B=0$

Let $2 x^{2}+x-3=t$

$\therefore(4 x+1) d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{4 x+1}{\sqrt{2 x^{2}+x-3}} d x & =\int \dfrac{1}{\sqrt{t}} d t \\ & =2 \sqrt{t}+C \\ & =2 \sqrt{2 x^{2}+x-3}+C \end{aligned} $

Solution

Let $x+2=A \dfrac{d}{d x}(x^{2}-1)+B$

$\Rightarrow x+2=A(2 x)+B \qquad …(1)$

Equating the coefficients of $x$ and constant term on both sides, we obtain

$2 A=1 \Rightarrow A=\dfrac{1}{2}$

$B=2$

From (1), we obtain

$(x+2)=\dfrac{1}{2}(2 x)+2$

Then, $\int \dfrac{x+2}{\sqrt{x^{2}-1}} d x=\int \dfrac{\dfrac{1}{2}(2 x)+2}{\sqrt{x^{2}-1}} d x$

$ \begin{aligned} =\dfrac{1}{2} \int \dfrac{2 x}{\sqrt{x^{2}-1}} d x+\int \dfrac{2}{\sqrt{x^{2}-1}} d x \qquad …(2) \end{aligned} $

In $\dfrac{1}{2} \int \dfrac{2 x}{\sqrt{x^{2}-1}} d x$, let $x^{2}-1=t \Rightarrow 2 x d x=d t$

$ \begin{aligned} \dfrac{1}{2} \int \dfrac{2 x}{\sqrt{x^{2}-1}} d x & =\dfrac{1}{2} \int \dfrac{d t}{\sqrt{t}} \\ & =\dfrac{1}{2}[2 \sqrt{t}] \\ & =\sqrt{t} \\ & =\sqrt{x^{2}-1} \end{aligned} $

Then, $\int \dfrac{2}{\sqrt{x^{2}-1}} d x=2 \int \dfrac{1}{\sqrt{x^{2}-1}} d x=2 \log |x+\sqrt{x^{2}-1}|$

From equation (2), we obtain

$ \int \dfrac{x+2}{\sqrt{x^{2}-1}} d x=\sqrt{x^{2}-1}+2 \log |x+\sqrt{x^{2}-1}|+C $

Solution

$\begin{aligned} &\int \dfrac{5x - 2}{1 + 2x + 3x^2} dx = I \quad (\text{say}) \ &\text{Let } 5x - 2 = A \cdot \dfrac{d}{dx} (1 + 2x + 3x^2) + B \ &\Rightarrow 5x - 2 = A (2 + 6x) + B \ &\text{Equating coefficients of } x, \ &5 = 6A \Rightarrow A = \dfrac{5}{6} \ &\text{Equating constant terms} \ &-2 = 2A + B \ &\Rightarrow -2 = 2 \times \dfrac{5}{6} + B \ &\Rightarrow B = -\dfrac{11}{3} \ &I = \dfrac{5}{6} \int \dfrac{2 + 6x}{1 + 2x + 3x^2} dx - \dfrac{11}{3} \int \dfrac{1}{1 + 2x + 3x^2} dx \ &I = \dfrac{5}{6} \int \dfrac{d(1 + 2x + 3x^2)}{1 + 2x + 3x^2} - \dfrac{11}{3} \int \dfrac{1}{1 + 2x + 3x^2} dx \ &I = \dfrac{5}{6} \log |1 + 2x + 3x^2| - \dfrac{11}{3} \int \dfrac{1}{1 + 2x + 3x^2} dx \ &\text{For first integral} \ &\text{Let } 1 + 2x + 3x^2 = t \ &\Rightarrow (2 + 6x) dx = dt \ &I = \dfrac{5}{6} \int \dfrac{dt}{t} - \dfrac{11}{3} \int \dfrac{1}{\sqrt{(x + \dfrac{1}{3})^2 + (\dfrac{\sqrt{2}}{3})^2}} dx \ &I = \dfrac{5}{6} \log |t| - \dfrac{11}{3} \left( \dfrac{1}{\dfrac{\sqrt{2}}{3}} \right) \tan^{-1} \left( \dfrac{x + \dfrac{1}{3}}{\dfrac{\sqrt{2}}{3}} \right) + c \ &I = \dfrac{5}{6} \log |1 + 2x + 3x^2| - \dfrac{11}{3} \cdot \dfrac{3}{\sqrt{2}} \tan^{-1} \left( \dfrac{3x + 1}{\sqrt{2}} \right) + c \ &I = \dfrac{5}{6} \log |1 + 2x + 3x^2| - \dfrac{11}{\sqrt{2}} \tan^{-1} \left( \dfrac{3x + 1}{\sqrt{2}} \right) + c \end{aligned} $

Solution

$\dfrac{6 x+7}{\sqrt{(x-5)(x-4)}}=\dfrac{6 x+7}{\sqrt{x^{2}-9 x+20}}$

Let $6 x+7=A \dfrac{d}{d x}(x^{2}-9 x+20)+B$

$\Rightarrow 6 x+7=A(2 x-9)+B$

Equating the coefficients of $x$ and constant term, we obtain

$2 A=6 \Rightarrow A=3$

$-9 A+B=7 \Rightarrow B=34$

$\therefore 6 x+7=3(2 x-9)+34$

$ \begin{aligned} \int \dfrac{6 x+7}{\sqrt{x^{2}-9 x+20}} & =\int \dfrac{3(2 x-9)+34}{\sqrt{x^{2}-9 x+20}} d x \\ & =3 \int \dfrac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x+34 \int \dfrac{1}{\sqrt{x^{2}-9 x+20}} d x \end{aligned} $

Let $I_1=\int \dfrac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x$ and $I_2=\int \dfrac{1}{\sqrt{x^{2}-9 x+20}} d x$

$\therefore \int \dfrac{6 x+7}{\sqrt{x^{2}-9 x+20}}=3 I_1+34 I_2 \qquad …(1)$

Then,

$I_1=\int \dfrac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x$

Let $x^{2}-9 x+20=t$

$\Rightarrow(2 x-9) d x=d t$

$\Rightarrow I_1=\dfrac{d t}{\sqrt{t}}$

$I_1=2 \sqrt{t}$

$I_1=2 \sqrt{x^{2}-9 x+20} \qquad …(2)$

and $I_2=\int \dfrac{1}{\sqrt{x^{2}-9 x+20}} d x$ $x^{2}-9 x+20$ can be written as $x^{2}-9 x+20+\dfrac{81}{4}-\dfrac{81}{4}$.

Therefore,

$x^{2}-9 x+20+\dfrac{81}{4}-\dfrac{81}{4}$

$=(x-\dfrac{9}{2})^{2}-\dfrac{1}{4}$

$=(x-\dfrac{9}{2})^{2}-(\dfrac{1}{2})^{2}$

$\Rightarrow I_2=\int \dfrac{1}{\sqrt{(x-\dfrac{9}{2})^{2}-(\dfrac{1}{2})^{2}}} d x$

$I_2=\log |(x-\dfrac{9}{2})+\sqrt{x^{2}-9 x+20}| \qquad …(3)$

Substituting equations (2) and (3) in (1), we obtain

$ \begin{aligned} \int \dfrac{6 x+7}{\sqrt{x^{2}-9 x+20}} d x & =3[2 \sqrt{x^{2}-9 x+20}]+34 \log [(x-\dfrac{9}{2})+\sqrt{x^{2}-9 x+20}]+C \\ & =6 \sqrt{x^{2}-9 x+20}+34 \log [(x-\dfrac{9}{2})+\sqrt{x^{2}-9 x+20}]+C \end{aligned} $

Solution

Let $x+2=A \dfrac{d}{d x}(4 x-x^{2})+B$

$\Rightarrow x+2=A(4-2 x)+B$

Equating the coefficients of $x$ and constant term on both sides, we obtain

$-2 A=1 \Rightarrow A=-\dfrac{1}{2}$

$4 A+B=2 \Rightarrow B=4$

$\Rightarrow(x+2)=-\dfrac{1}{2}(4-2 x)+4$

$\therefore \int \dfrac{x+2}{\sqrt{4 x-x^{2}}} d x=\int \dfrac{-\dfrac{1}{2}(4-2 x)+4}{\sqrt{4 x-x^{2}}} d x$ $=-\dfrac{1}{2} \int \dfrac{4-2 x}{\sqrt{4 x-x^{2}}} d x+4 \int \dfrac{1}{\sqrt{4 x-x^{2}}} d x$

Let $I_1=\int \dfrac{4-2 x}{\sqrt{4 x-x^{2}}} d x$ and $I_2= \int \dfrac{1}{\sqrt{4 x-x^{2}}} d x$

$\therefore \int \dfrac{x+2}{\sqrt{4 x-x^{2}}} d x=-\dfrac{1}{2} I_1+4 I_2 \qquad …(1)$

Then, $I_1=\int \dfrac{4-2 x}{\sqrt{4 x-x^{2}}} d x$

Let $4 x-x^{2}=t$

$\Rightarrow(4-2 x) d x=d t$

$\Rightarrow I_1=\int{\dfrac {d t} {\sqrt{t}}}=2 \sqrt{t}=2 \sqrt{4 x-x^{2}} \qquad …(2)$

$I_2=\int \dfrac{1}{\sqrt{4 x-x^{2}}} d x$

$\Rightarrow 4 x-x^{2}=-(-4 x+x^{2})$

$=(-4 x+x^{2}+4-4)$

$=4-(x-2)^{2}$

$=(2)^{2}-(x-2)^{2}$

$\therefore I_2=\int \dfrac{1}{\sqrt{(2)^{2}-(x-2)^{2}}} d x=\sin ^{-1}(\dfrac{x-2}{2}) \qquad …(3)$

Using equations (2) and (3) in (1), we obtain

$ \begin{aligned} \int \dfrac{x+2}{\sqrt{4 x-x^{2}}} d x & =-\dfrac{1}{2}(2 \sqrt{4 x-x^{2}})+4 \sin ^{-1}(\dfrac{x-2}{2})+C \\ & =-\sqrt{4 x-x^{2}}+4 \sin ^{-1}(\dfrac{x-2}{2})+C \end{aligned} $

Solution

$\begin{aligned} &\int \dfrac{(x+2)}{\sqrt{x^2+2x+3}} , dx = \dfrac{1}{2} \int \dfrac{2(x+2)}{\sqrt{x^2+2x+3}} , dx \ &= \dfrac{1}{2} \int \dfrac{2x+4}{\sqrt{x^2+2x+3}} , dx \ &= \dfrac{1}{2} \int \dfrac{2x+2+2}{\sqrt{x^2+2x+3}} , dx \ &= \dfrac{1}{2} \int \dfrac{2x+2}{\sqrt{x^2+2x+3}} , dx + \dfrac{1}{2} \int \dfrac{2}{\sqrt{x^2+2x+3}} , dx \ &= \dfrac{1}{2} \int \dfrac{2x+2}{\sqrt{x^2+2x+3}} , dx + \int \dfrac{1}{\sqrt{x^2+2x+3}} , dx \ &\text{Let } I_1 = \int \dfrac{2x+2}{\sqrt{x^2+2x+3}} , dx \text{ and } I_2 = \int \dfrac{1}{\sqrt{x^2+2x+3}} , dx \ &\therefore \int \dfrac{x+2}{\sqrt{x^2+2x+3}} , dx = \dfrac{1}{2} I_1 + I_2 \quad \text{……..(1)} \ &\text{Then, } I_1 = \int \dfrac{2x+2}{\sqrt{x^2+2x+3}} , dx \ &\text{Let } x^2 + 2x + 3 = t \ &\Rightarrow (2x+2) , dx = dt \ &I_1 = \int \dfrac{dt}{\sqrt{t}} = 2\sqrt{t} = 2\sqrt{x^2+2x+3} \quad \text{……..(2)} \ &I_2 = \int \dfrac{1}{\sqrt{x^2+2x+3}} , dx \ &\Rightarrow x^2 + 2x + 3 = (x+1)^2 + (\sqrt{2})^2 \ &\Rightarrow I_2 = \int \dfrac{1}{\sqrt{(x+1)^2 + (\sqrt{2})^2}} , dx = \log \left| (x+1) + \sqrt{x^2+2x+3} \right| \quad \text{……..(3)} \ &\text{Using equation (2) and (3) in (1), we obtain} \ &\int \dfrac{x+2}{\sqrt{x^2+2x+3}} , dx = \dfrac{1}{2} \left[ 2\sqrt{x^2+2x+3} \right] + \log \left| (x+1) + \sqrt{x^2+2x+3} \right| + C \ &= \sqrt{x^2+2x+3} + \log \left| (x+1) + \sqrt{x^2+2x+3} \right| + C \end{aligned} $

Solution

Let $(x+3)=A \dfrac{d}{d x}(x^{2}-2 x-5)+B$

$(x+3)=A(2 x-2)+B$

Equating the coefficients of $x$ and constant term on both sides, we obtain

$ \begin{aligned} & 2 A=1 \Rightarrow A=\dfrac{1}{2} \\ & -2 A+B=3 \Rightarrow B=4 \\ & \therefore(x+3)=\dfrac{1}{2}(2 x-2)+4 \\ & \Rightarrow \int \dfrac{x+3}{x^{2}-2 x-5} d x=\int \dfrac{\dfrac{1}{2}(2 x-2)+4}{x^{2}-2 x-5} d x \\ & \quad=\dfrac{1}{2} \int \dfrac{2 x-2}{x^{2}-2 x-5} d x+4 \int \dfrac{1}{x^{2}-2 x-5} d x \end{aligned} $

Let $I_1=\int \dfrac{2 x-2}{x^{2}-2 x-5} d x$ and $I_2=\int \dfrac{1}{x^{2}-2 x-5} d x$

$\therefore \int \dfrac{x+3}{(x^{2}-2 x-5)} d x=\dfrac{1}{2} I_1+4 I_2 \qquad …(1)$

Then, $I_1=\int \dfrac{2 x-2}{x^{2}-2 x-5} d x$

Let $x^{2}-2 x-5=t$

$\Rightarrow(2 x-2) d x=d t$

$\Rightarrow I_1=\int \dfrac{d t}{t}=\log |t|=\log |x^{2}-2 x-5| \qquad …(2)$

$I_2=\int \dfrac{1}{x^{2}-2 x-5} d x$

$=\int \dfrac{1}{(x^{2}-2 x+1)-6} d x$

$=\int \dfrac{1}{(x-1)^{2}+(\sqrt{6})^{2}} d x$

$=\dfrac{1}{2 \sqrt{6}} \log (\dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}}) \qquad …(3)$

Substituting (2) and (3) in (1), we obtain

$ \begin{aligned} \int \dfrac{x+3}{x^{2}-2 x-5} d x & =\dfrac{1}{2} \log |x^{2}-2 x-5|+\dfrac{4}{2 \sqrt{6}} \log |\dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C \\ & =\dfrac{1}{2} \log |x^{2}-2 x-5|+\dfrac{2}{\sqrt{6}} \log |\dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C \end{aligned} $

Solution

Let $5 x+3=A \dfrac{d}{d x}(x^{2}+4 x+10)+B$

$\Rightarrow 5 x+3=A(2 x+4)+B$

Equating the coefficients of $x$ and constant term, we obtain

$ \begin{aligned} & 2 A=5 \Rightarrow A=\dfrac{5}{2} \\ & 4 A+B=3 \Rightarrow B=-7 \\ & \therefore 5 x+3=\dfrac{5}{2}(2 x+4)-7 \\ & \Rightarrow \int \dfrac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\int \dfrac{\dfrac{5}{2}(2 x+4)-7}{\sqrt{x^{2}+4 x+10}} d x \\ & \quad=\dfrac{5}{2} \int \dfrac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \dfrac{1}{\sqrt{x^{2}+4 x+10}} d x \end{aligned} $

Let $I_1=\int \dfrac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$ and $I_2=\int \dfrac{1}{\sqrt{x^{2}+4 x+10}} d x$

$\therefore \int \dfrac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\dfrac{5}{2} I_1-7 I_2 \qquad …(1)$

Then, $I_1=\int \dfrac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$

Let $x^{2}+4 x+10=t$

$\therefore(2 x+4) d x=d t$

$\Rightarrow I_1=\int \dfrac{d t}{t}=2 \sqrt{t}=2 \sqrt{x^{2}+4 x+10} \qquad …(2)$

$I_2=\int \dfrac{1}{\sqrt{x^{2}+4 x+10}} d x$

$=\int \dfrac{1}{\sqrt{(x^{2}+4 x+4)+6}} d x$

$=\int \dfrac{1}{(x+2)^{2}+(\sqrt{6})^{2}} d x$

$=\log |(x+2) \sqrt{x^{2}+4 x+10}| \qquad …(3)$

Using equations (2) and (3) in (1), we obtain

$ \begin{aligned} \int \dfrac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x & =\dfrac{5}{2}[2 \sqrt{x^{2}+4 x+10}]-7 \log |(x+2)+\sqrt{x^{2}+4 x+10}|+C \\ & =5 \sqrt{x^{2}+4 x+10}-7 \log (x+2)+\sqrt{x^{2}+4 x+10} \mid+C \end{aligned} $

Solution

$ \begin{aligned} \int \dfrac{d x}{x^{2}+2 x+2} & =\int \dfrac{d x}{(x^{2}+2 x+1)+1} \\ & =\int \dfrac{1}{(x+1)^{2}+(1)^{2}} d x \\ & =[\tan ^{-1}(x+1)]+C \end{aligned} $

Hence, the correct Answer is B.

Solution

$\int \dfrac{d x}{\sqrt{9 x-4 x^{2}}}$

$=\int \dfrac{1}{\sqrt{-4(x^{2}-\dfrac{9}{4} x)}} d x$

$=\int \dfrac{1}{-4(x^{2}-\dfrac{9}{4} x+\dfrac{81}{64}-\dfrac{81}{64})} d x$

$=\int \dfrac{1}{\sqrt{-4[(x-\dfrac{9}{8})^{2}-(\dfrac{9}{8})^{2}]}} d x$

$=\dfrac{1}{2} \int \dfrac{1}{\sqrt{(\dfrac{9}{8})^{2}-(x-\dfrac{9}{8})^{2}}} d x$

$=\dfrac{1}{2}[\sin ^{-1}(\dfrac{x-\dfrac{9}{8}}{\dfrac{9}{8}})]+C\qquad (\because \int \dfrac{d y}{\sqrt{a^{2}-y^{2}}}=\sin ^{-1} \dfrac{y}{a}+C)$

$=\dfrac{1}{2} \sin ^{-1}(\dfrac{8 x-9}{9})+C$

Hence, the correct Answer is B.