Question

perform the integration, integrate dx/(1-x) √1+x)​

Answer 1

Given Integrand,

$\displaystyle \sf \int \dfrac{dx}{(x - 1) \sqrt{x + 1} }$

Let y² = x + 1.

Differentiating w.r.t x,

$\longrightarrow \sf \: 2y \dfrac{dy}{dx} = 1 \\ \\ \longrightarrow \sf 2ydy = dx$

Also, x = y² - 1.

Now,

$\longrightarrow \displaystyle \sf \int \dfrac{2ydy}{(y {}^{2} - 1 - 1) y} \\ \\ \longrightarrow \displaystyle \sf 2\int \dfrac{dy}{(y {}^{2} - 2)}$

We know that,

$\displaystyle \star \: \boxed { \boxed{\sf \int \dfrac{dx}{ {x}^{2} - {a}^{2} } = \frac{1}{2a} log \bigg| \dfrac{x - a}{x + a} \bigg| + C}}$

Now,

$\longrightarrow \sf \: 2 \times \dfrac{1}{2 \times 2} log \bigg| \dfrac{y - 2}{y + 2} \bigg| + C \\ \\ \longrightarrow \sf \dfrac{1}{2} log \bigg| \dfrac{x + 1 - 2}{x + 1 + 2} \bigg| + C \\ \\ \longrightarrow \sf \dfrac{1}{2} log \bigg| \dfrac{x - 1}{x + 3} \bigg| + C$

Thus,

$\star \: \displaystyle \boxed{ \boxed{ \sf \int \dfrac{dx}{(x - 1) \sqrt{x + 1} } = \dfrac{1}{2} log \bigg| \dfrac{x - 1}{x + 3} \bigg| + C}}$