Question

$\int \frac{1}{(x+1{)}^{3/2}(x-1{)}^{1/2}}dx=$

• A. $-{\sin }^{-1}\left[\frac{1-x}{(x+2)\sqrt{2}}\right]+c$
• B. $\sqrt{\frac{x+1}{x-1}}+c$
• C. $\sqrt{1-\frac{2+x}{x+1}}+c$
• D. $\sqrt{\frac{x-1}{x+1}}+c$

Answer 1

The correct option is D $\sqrt{\frac{x-1}{x+1}}+c$

$\int \frac{1}{(x+1{)}^{3{/}_2}(x-1{)}^1{/}_2}dx$

$\int \frac{1}{x^{3/2}\cdot x^{1/2}\left(1{+}^1{/}_x{)}^{3/2}\right(1{-}^1/x)}dx$

$=\int \frac{dx}{x^2\left(1{+}^1{/}_x{)}^{3/2}\right(1{-}^1{/}_x{)}^{1/2}}$

$\frac{1}{x}=t$

${}^{-1}{/}_{x^2}dx=dt{\Rightarrow }^1{/}_{x^2}dx=-dt\int \frac{-dt}{(1+t{)}^{3/2}(1-t{)}^{1/2}\cdot }$

$t=cos2\theta$

$dt=-2sin2\theta \cdot d\theta$

$=\int \frac{2sin2\theta d\theta }{\left(2co{s}^2\theta {)}^{3/2}\right(2si{n}^2\theta {)}^{1/2}}$

$=\int \frac{2_{1}sin\theta cos\theta }{4co{s}^2\theta \sin \theta }d\theta$

$=\int se{c}^{2}d\theta =tan\theta +c\cdot$

$=\sqrt{\frac{1-cos2\theta }{1+co{s}^2\theta }}+c$

$\sqrt{\frac{1{-}^1{/}_x\cdot }{1{+}^1{/}_x}}+c$

$=\sqrt{\frac{x-1}{x+1}}+c\cdot$