Integral Calculus Question 47
Question: $ \int{\sqrt{x^{2}+a^{2}}dx} $ equals to
[RPET 2001]
Options:
A) $ \frac{x}{2}\sqrt{x^{2}+a^{2}}-\frac{a^{2}}{2} \log | {x+\sqrt{x^{2}+a^{2}}}|+c $
B) $ \frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log |{x+\sqrt{x^{2}+a^{2}}}|+c $
C) $ \frac{x}{2}\sqrt{x^{2}+a^{2}}-\frac{a^{2}}{2} \log |{x-\sqrt{x^{2}+a^{2}}}|+c $
D) $ \frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log |{x-\sqrt{x^{2}+a^{2}}}|+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int{\sqrt{x^{2}+a^{2}}dx} $ $ =\int{\sqrt{x^{2}+a^{2}}.1dx} $
$ =\sqrt{x^{2}+a^{2}}\int{1dx-\int{[ \frac{d}{dx}( \sqrt{x^{2}+a^{2}} )\int{1dx} ]dx}} $
$ =x\sqrt{x^{2}+a^{2}}-\int{[ \frac{2x}{2\sqrt{x^{2}+a^{2}}}x ]}dx $
$ =x\sqrt{x^{2}+a^{2}}-\int{[ \frac{x^{2}+a^{2}-a^{2}}{\sqrt{x^{2}+a^{2}}} ]}dx $
$ =x\sqrt{x^{2}+a^{2}}-\int{[ \sqrt{x^{2}+a^{2}}-\frac{a^{2}}{\sqrt{x^{2}+a^{2}}} ]}dx $
$ =x\sqrt{x^{2}+a^{2}}-\int{\sqrt{x^{2}+a^{2}}dx+a^{2}\int{\frac{dx}{\sqrt{x^{2}+a^{2}}}}} $
$ =\sqrt{x^{2}+a^{2}}-I+a^{2}\log [ x+\sqrt{x^{2}+a^{2}} ]+C $
$ 2I=x\sqrt{x^{2}+a^{2}+}a^{2}\log [ x+\sqrt{x^{2}+a^{2}} ]+C $
$ I=\frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2}\log [ x+\sqrt{x^{2}+a^{2}} ]+C $ .
BETA
