Question

$\int (\sqrt{\frac{a+x}{a-x}}-\sqrt{\frac{a-x}{a+x}})dx$ is equal to

• A. $-2\sqrt{a^2-x^2}+c$
• B. $2\sqrt{a^2-x^2}+c$
• C. $-2\sqrt{x^2-a^2}+c$
• D. $2\sqrt{x^2-a^2}+c$

Answer 1

The correct option is A $-2\sqrt{a^2-x^2}+c$

Let $I=\int (\sqrt{\frac{a+x}{a-x}}-\sqrt{\frac{a-x}{a+x}})dx$

$I=\int \frac{((a+x)-(a-x))}{\sqrt{(a-x)(a+x)}}dx$

$I=\int \frac{2x}{\sqrt{(a-x)(a+x)}}dx$

$I=\int \frac{2x}{\sqrt{(a^2-x^2)}}.dx$

$N\left(x\right)=2x$ (Numerator)

$D\left(x\right)=a^2-x^2$ (Denominator)

$D^{\prime }\left(x\right)=-2x$

$N\left(x\right)=-D^{\prime }\left(x\right)$

Thus, $I=-\frac{\int D^{\prime }\left(x\right)d\left(x\right)}{\sqrt{D\left(x\right)}}$

$I=-2\sqrt{D\left(x\right)}+c$

$I=-2\sqrt{(a^2-x^2)}+c$

$I=-2\sqrt{(a^2-x^2)}+c$