Question

Evaluate: $\int \sqrt{\frac {1+x}{1-x}} \ dx ,x \neq 1$

Answer 1

$\underline{\text{Solution :}}$

$\mathrm{Now,\:\int \sqrt{\frac{1+x}{1-x}}\:dx}$

$\mathrm{=\int \sqrt{\frac{(1+x)(1+x)}{(1-x)(1+x)}}\:dx}$

$\mathrm{=\int \sqrt{\frac{(1+x)^{2}}{1-x^{2}}}\:dx}$

$\mathrm{=\int \frac{1+x}{\sqrt{1-x^{2}}}\:dx}$

$\mathrm{=\int \frac{dx}{\sqrt{1-x^{2}}}+\int \frac{x\:dx}{\sqrt{1-x^{2}}}}$

$\mathrm{=sin^{-1}x+\sqrt{1-x^{2}}+C}$

$\text{where C is integral constant}$

$\to \boxed{\mathrm{\int \sqrt{\frac{1+x}{1-x}}\:dx=sin^{-1}x+\sqrt{1-x^{2}}+C}}$

$\text{which is the required integral}$

$\star\:\underline{\text{Method}}\:\star$

$\mathrm{Let,\:\sqrt{1-x^{2}}=z}$

$\to \mathrm{-\frac{1}{2}\frac{-2x}{\sqrt{1-x^{2}}}\:dx=dz}$

$\to \mathrm{\frac{x\:dx}{\sqrt{1-x^{2}}}=dz}$

$\to \mathrm{\int \frac{x\:dx}{\sqrt{1-x^{2}}}=\int dz}$

$\to \mathrm{\frac{x\:dx}{\sqrt{1-x^{2}}}=z+C}$

Answer 2

ANSWER:--------------

∫ 1−x1+xdx

=∫(1+x)(1+x)(1−x)(1+x)dx\mathrm{=\int \sqrt{\frac{(1+x)(1+x)}{(1-

x)(1+x)}}\:dx}=∫

(1−x)(1+x)

(1+x)(1+x)dx

=∫(1+x)21−x2dx\mathrm{=\int \sqrt{\frac{(1+x)^{2}}{1-x^{2}}}\:dx}

=∫ 1−x 2(1+x) 2dx

=∫1+x1−x2dx\mathrm{=\int \frac{1+x}{\sqrt{1-x^{2}}}\:dx}

=∫ 1−x 21+xdx

=∫dx1−x2+∫xdx1−x2\mathrm{=\int \frac{dx}{\sqrt{1-x^{2}}}+\int \frac{x\:dx}{\sqrt{1-x^{2}}}}=∫

1−x 2dx

+∫ 1−x 2xdx

=sin−1x+1−x2+C\mathrm

{=sin^{-1}x+\sqrt{1-x^{2}}+C}=sin

−1x+ 1−x 2+C

C integral constant

\text{where C is constant}

{C is integral constant }

→∫1+x1−xdx

=sin−1x+1−x2+C\to

\boxed{\mathrm

{\int \sqrt{\frac{1+x}

{1-x}}\:dx=sin^{-1}x+\sqrt

{1-x^{2}}+C}}→

∫ 1−x1+x

dx=sin −1x+ 1−x 2+C

hope it helps:---

T!—!ANKS!!!