Solution

Let $I=\int _{-1}^{1}(x+1) d x$

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

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(-1) \\ & =(\dfrac{1}{2}+1)-(\dfrac{1}{2}-1) \\ & =\dfrac{1}{2}+1-\dfrac{1}{2}+1 \\ & =2 \end{aligned} $

Solution

Let $I=\int_2^{3} \dfrac{1}{x} d x$

$\dfrac{1}{x} d x=\log |x|=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(3)-F(2) \\ & =\log |3|-\log |2|=\log \dfrac{3}{2} \end{aligned} $

Solution

Let $I=\int_1^{2}(4 x^{3}-5 x^{2}+6 x+9) d x$

$\int(4 x^{3}-5 x^{2}+6 x+9) d x=4(\dfrac{x^{4}}{4})-5(\dfrac{x^{3}}{3})+6(\dfrac{x^{2}}{2})+9(x)$

$=x^{4}-\dfrac{5 x^{3}}{3}+3 x^{2}+9 x=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(2)-F(1) \\ I & ={2^{4}-\dfrac{5 \cdot(2)^{3}}{3}+3(2)^{2}+9(2)}-{(1)^{4}-\dfrac{5(1)^{3}}{3}+3(1)^{2}+9(1)} \\ & =(16-\dfrac{40}{3}+12+18)-(1-\dfrac{5}{3}+3+9) \\ & =16-\dfrac{40}{3}+12+18-1+\dfrac{5}{3}-3-9 \\ & =33-\dfrac{35}{3} \\ & =\dfrac{99-35}{3} \\ & =\dfrac{64}{3} \end{aligned} $

Solution

Let $I=\int_0^{\pi} \sin 2 x d x$

$\int \sin 2 x d x=(\dfrac{-\cos 2 x}{2})=F(x)$

By second fundamental theorem of calculus, we obtain

$ I =F(\dfrac{\pi}{4})-F(0) $

$ =-\dfrac{1}{2}[\cos 2(\dfrac{\pi}{4})-\cos 0] $

$ =-\dfrac{1}{2}[\cos (\mathrm{\pi} / 2)-1] $

$ =-\dfrac{1}{2}[0-1] $

$ =\dfrac{1}{2} $

Solution

Let $I=\int_0^{\dfrac{\pi}{2}} \cos 2 x d x$

$\int \cos 2 x d x=(\dfrac{\sin 2 x}{2})=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(\dfrac{\pi}{2})-F(0) \\ & =\dfrac{1}{2}[\sin 2(\dfrac{\pi}{2})-\sin 0] \\ & =\dfrac{1}{2}[\sin \pi-\sin 0] \\ & =\dfrac{1}{2}[0-0]=0 \end{aligned} $

Solution

Let $I=\int_4^{5} e^{x} d x$

$\int e^{x} d x=e^{x}=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(5)-F(4) \\ & =e^{5}-e^{4} \\ & =e^{4}(e-1) \end{aligned} $

Solution

Let $I=\int_0^{\dfrac{\pi}{4}} \tan x d x$

$\int \tan x d x=-\log |\cos x|=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(\dfrac{\pi}{4})-F(0) \\ & =-\log |\cos \dfrac{\pi}{4}|+\log |\cos 0| \\ & =-\log |\dfrac{1}{\sqrt{2}}|+\log |1| \\ & =-\log (2)^{-\dfrac{1}{2}} \\ & =\dfrac{1}{2} \log 2 \end{aligned} $

Solution

Let $I=\int _{\dfrac{\pi}{6}}^{\dfrac{\pi}{4}} cosec x d x$

$\int cosec x d x=\log |cosec x-\cot x|=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(\dfrac{\pi}{4})-F(\dfrac{\pi}{6}) \\ & =\log |cosec \dfrac{\pi}{4}-\cot \dfrac{\pi}{4}|-\log |cosec \dfrac{\pi}{6}-\cot \dfrac{\pi}{6}| \\ & =\log |\sqrt{2}-1|-\log |2-\sqrt{3}| \\ & =\log (\dfrac{\sqrt{2}-1}{2-\sqrt{3}}) \end{aligned} $

Solution

Let $I=\int_0^{1} \dfrac{d x}{\sqrt{1-x^{2}}}$

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

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & =\sin ^{-1}(1)-\sin ^{-1}(0) \\ & =\dfrac{\pi}{2}-0 \\ & =\dfrac{\pi}{2} \end{aligned} $

Solution

Let $I=\int_0^{1} \dfrac{d x}{1+x^{2}}$

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

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & =\tan ^{-1}(1)-\tan ^{-1}(0) \\ & =\dfrac{\pi}{4} \end{aligned} $

Solution

Let $I=\int_2^{3} \dfrac{d x}{x^{2}-1}$

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

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(3)-F(2) \\ & =\dfrac{1}{2}[\log |\dfrac{3-1}{3+1}|-\log |\dfrac{2-1}{2+1}|] \\ & =\dfrac{1}{2}[\log |\dfrac{2}{4}|-\log |\dfrac{1}{3}|] \\ & =\dfrac{1}{2}[\log \dfrac{1}{2}-\log \dfrac{1}{3}] \\ & =\dfrac{1}{2}[\log \dfrac{3}{2}] \end{aligned} $

Solution

Let $I=\int_0^{\dfrac{\pi}{2}} \cos ^{2} x d x$

$\int \cos ^{2} x d x=\int(\dfrac{1+\cos 2 x}{2}) d x=\dfrac{x}{2}+\dfrac{\sin 2 x}{4}=\dfrac{1}{2}(x+\dfrac{\sin 2 x}{2})=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =[F(\dfrac{\pi}{2})-F(0)] \\ & =\dfrac{1}{2}[(\dfrac{\pi}{2}-\dfrac{\sin \pi}{2})-(0+\dfrac{\sin 0}{2})] \\ & =\dfrac{1}{2}[\dfrac{\pi}{2}+0-0-0] \\ & =\dfrac{\pi}{4} \end{aligned} $

Solution

Let $I=\int_2^{3} \dfrac{x}{x^{2}+1} d x$

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

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(3)-F(2) \\ & =\dfrac{1}{2}[\log (1+(3)^{2})-\log (1+(2)^{2})] \\ & =\dfrac{1}{2}[\log (10)-\log (5)] \\ & =\dfrac{1}{2} \log (\dfrac{10}{5})=\dfrac{1}{2} \log 2 \end{aligned} $

Solution

Let $I=\int_0^{1} \dfrac{2 x+3}{5 x^{2}+1} d x$

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

$=\dfrac{1}{5} \int \dfrac{10 x+15}{5 x^{2}+1} d x$

$=\dfrac{1}{5} \int \dfrac{10 x}{5 x^{2}+1} d x+3 \int \dfrac{1}{5 x^{2}+1} d x$

$=\dfrac{1}{5} \int \dfrac{10 x}{5 x^{2}+1} d x+3 \int \dfrac{1}{5(x^{2}+\dfrac{1}{5})} d x$

$=\dfrac{1}{5} \log (5 x^{2}+1)+\dfrac{3}{5} \cdot \dfrac{1}{\dfrac{1}{\sqrt{5}}} \tan ^{-1} \dfrac{x}{\dfrac{1}{\sqrt{5}}}$

$=\dfrac{1}{5} \log (5 x^{2}+1)+\dfrac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5} x)$

$=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & ={\dfrac{1}{5} \log (5+1)+\dfrac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5})}-{\dfrac{1}{5} \log (1)+\dfrac{3}{\sqrt{5}} \tan ^{-1}(0)} \\ & =\dfrac{1}{5} \log 6+\dfrac{3}{\sqrt{5}} \tan ^{-1} \sqrt{5} \end{aligned} $

Solution

Let $I=\int_0^{1} x e^{x^{2}} d x$

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

As $x \to 0, t \to 0$ and as $x \to 1, t \to 1$,

$\therefore I=\dfrac{1}{2} \int_0^{1} e^{t} d t$

$\dfrac{1}{2} \int e^{t} d t=\dfrac{1}{2} e^{t}=F(t)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & =\dfrac{1}{2} e-\dfrac{1}{2} e^{0} \\ & =\dfrac{1}{2}(e-1) \end{aligned} $

Solution

Let $I=\int_1^{2} \dfrac{5 x^{2}}{x^{2}+4 x+3} d x$

Dividing $5 x^{2}$ by $x^{2}+4 x+3$, we obtain

$I=\int_1^{2}{5-\dfrac{20 x+15}{x^{2}+4 x+3}} d x$

$=\int_1^{2} 5 d x-\int_1^{2} \dfrac{20 x+15}{x^{2}+4 x+3} d x$

$=[5 x]_1^{2}-\int_1^{2} \dfrac{20 x+15}{x^{2}+4 x+3} d x$

$I=5-I_1$, where $I_1=\int_1^{2} \dfrac{20 x+15}{x^{2}+4 x+3} d x \qquad …(1)$

Let $20 x+15=A \dfrac{d}{d x}(x^{2}+4 x+3)+B$

$ =2 A x+(4 A+B) $

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

$A=10$ and $B=-25$

$\Rightarrow I_1=10 \int_1^{2} \dfrac{2 x+4}{x^{2}+4 x+3} d x-25 \int_1^{2} \dfrac{d x}{x^{2}+4 x+3}$

Let $x^{2}+4 x+3=t$

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

$\Rightarrow I_1=10 \int \dfrac{d t}{t}-25 \int \dfrac{d x}{(x+2)^{2}-1^{2}}$

$=10 \log t-25[\dfrac{1}{2} \log (\dfrac{x+2-1}{x+2+1})]$

$=[10 \log (x^{2}+4 x+3)]_1^{2}-25[\dfrac{1}{2} \log (\dfrac{x+1}{x+3})]_1^{2}$

$=[10 \log 15-10 \log 8]-25[\dfrac{1}{2} \log \dfrac{3}{5}-\dfrac{1}{2} \log \dfrac{2}{4}]$

$=[10 \log (5 \times 3)-10 \log (4 \times 2)]-\dfrac{25}{2}[\log 3-\log 5-\log 2+\log 4]$

$=[10 \log 5+10 \log 3-10 \log 4-10 \log 2]-\dfrac{25}{2}[\log 3-\log 5-\log 2+\log 4]$

$=[10+\dfrac{25}{2}] \log 5+[-10-\dfrac{25}{2}] \log 4+[10-\dfrac{25}{2}] \log 3+[-10+\dfrac{25}{2}] \log 2$

$=\dfrac{45}{2} \log 5-\dfrac{45}{2} \log 4-\dfrac{5}{2} \log 3+\dfrac{5}{2} \log 2$

$=\dfrac{45}{2} \log \dfrac{5}{4}-\dfrac{5}{2} \log \dfrac{3}{2}$

Substituting the value of $I_1$ in (1), we obtain

$ \begin{aligned} I & =5-[\dfrac{45}{2} \log \dfrac{5}{4}-\dfrac{5}{2} \log \dfrac{3}{2}] \\ & =5-\dfrac{5}{2}[9 \log \dfrac{5}{4}-\log \dfrac{3}{2}] \end{aligned} $

Solution

Let $I=\int_0^{\dfrac{\pi}{4}}(2 \sec ^{2} x+x^{3}+2) d x$

$\int(2 \sec ^{2} x+x^{3}+2) d x=2 \tan x+\dfrac{x^{4}}{4}+2 x=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(\dfrac{\pi}{4})-F(0) \\ & ={(2 \tan \dfrac{\pi}{4}+\dfrac{1}{4}(\dfrac{\pi}{4})^{4}+2(\dfrac{\pi}{4}))-(2 \tan 0+0+0)} \\ & =2 \tan \dfrac{\pi}{4}+\dfrac{\pi^{4}}{4^{5}}+\dfrac{\pi}{2} \\ & =2+\dfrac{\pi}{2}+\dfrac{\pi^{4}}{1024} \end{aligned} $

Solution

Let $I=\int_0^{\pi}(\sin ^{2} \dfrac{x}{2}-\cos ^{2} \dfrac{x}{2}) d x$

$ \begin{aligned} & =-\int_0^{\pi}(\cos ^{2} \dfrac{x}{2}-\sin ^{2} \dfrac{x}{2}) d x \\ & =-\int_0^{\pi} \cos x d x \end{aligned} $

$\int \cos x d x=\sin x=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(\pi)-F(0) \\ & =\sin \pi-\sin 0 \\ & =0 \end{aligned} $

Solution

Let $I=\int_0^{2} \dfrac{6 x+3}{x^{2}+4} d x$

$ \begin{aligned} \int \dfrac{6 x+3}{x^{2}+4} d x & =3 \int \dfrac{2 x+1}{x^{2}+4} d x \\ & =3 \int \dfrac{2 x}{x^{2}+4} d x+3 \int \dfrac{1}{x^{2}+4} d x \\ & =3 \log (x^{2}+4)+\dfrac{3}{2} \tan ^{-1} \dfrac{x}{2}=F(x) \end{aligned} $

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(2)-F(0) \\ & ={3 \log (2^{2}+4)+\dfrac{3}{2} \tan ^{-1}(\dfrac{2}{2})}-{3 \log (0+4)+\dfrac{3}{2} \tan ^{-1}(\dfrac{0}{2})} \\ & =3 \log 8+\dfrac{3}{2} \tan ^{-1} 1-3 \log 4-\dfrac{3}{2} \tan ^{-1} 0 \\ & =3 \log 8+\dfrac{3}{2}(\dfrac{\pi}{4})-3 \log 4-0 \\ & =3 \log (\dfrac{8}{4})+\dfrac{3 \pi}{8} \\ & =3 \log 2+\dfrac{3 \pi}{8} \end{aligned} $

Solution

Let $I=\int_0^{1}(x e^{x}+\sin \dfrac{\pi x}{4}) d x$

$ \begin{aligned} \int(x e^{x}+\sin \dfrac{\pi x}{4}) d x & =x \int e^{x} d x-\int{\dfrac{d}{d x}( x) \int e^{x} d x} d x+{\dfrac{-\cos \dfrac{\pi x}{4}}{\dfrac{\pi}{4}}} \\ & =x e^{x}-\int e^{x} d x-\dfrac{4 cos(\dfrac{\pi x}{4})}{\pi} \\ & =x e^{x}-e^{x}-\dfrac{4 cos(\dfrac{\pi x}{4})}{\pi} \\ & =F(x) \end{aligned} $

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & =(1 . e^{1}-e^{1}-\dfrac{4}{\pi} \cos \dfrac{\pi}{4})-(0 . e^{0}-e^{0}-\dfrac{4}{\pi} \cos 0) \\ & =e-e-\dfrac{4}{\pi}(\dfrac{1}{\sqrt{2}})+1+\dfrac{4}{\pi} \\ & =1+\dfrac{4}{\pi}-\dfrac{2 \sqrt{2}}{\pi} \end{aligned} $

Solution

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

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} & \int_1^{\sqrt{3}} \dfrac{d x}{1+x^{2}}=F(\sqrt{3})-F(1) \\ & =\tan ^{-1} \sqrt{3}-\tan ^{-1} 1 \\ & =\dfrac{\pi}{3}-\dfrac{\pi}{4} \\ & =\dfrac{\pi}{12} \end{aligned} $

Hence, the correct Answer is D.

Solution

$ \int \dfrac{d x}{4+9 x^{2}}=\int \dfrac{d x}{(2)^{2}+(3 x)^{2}} $

Put $3 x=t \Rightarrow 3 d x=d t$

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

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

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} \int_0^{\dfrac{2}{3}} \dfrac{d x}{4+9 x^{2}} & =F(\dfrac{2}{3})-F(0) \\ & =\dfrac{1}{6} \tan ^{-1}(\dfrac{3}{2} \cdot \dfrac{2}{3})-\dfrac{1}{6} \tan ^{-1} 0 \\ & =\dfrac{1}{6} \tan ^{-1} 1-0 \\ & =\dfrac{1}{6} \times \dfrac{\pi}{4} \\ & =\dfrac{\pi}{24} \end{aligned} $

Hence, the correct Answer is C.