int frac{1}{(x - 1)(x^2 + 1)} dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We use partial fractions: $\frac{1}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$.$1 = A(x^2 + 1) + (Bx + C)(x - 1)$.Setting $x =

Vedclass · Accepted Answer

$\frac{1}{2}\log |x - 1| - \frac{1}{4}\log (x^2 + 1) - \frac{1}{2}	an^{-1} x + c$

Vedclass · Answer

(A) We use partial fractions: $\frac{1}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$.$1 = A(x^2 + 1) + (Bx + C)(x - 1)$.Setting $x = 1$,we get $1 = A(1 + 1) \Rightarrow A = \frac{1}{2}$.Comparing coefficients of $x^2$: $A + B = 0 \Rightarrow B = -\frac{1}{2}$.Comparing constant terms: $A - C = 1 \Rightarrow C = A - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$.Thus,$\frac{1}{(x - 1)(x^2 + 1)} = \frac{1}{2(x - 1)} - \frac{x + 1}{2(x^2 + 1)}$.Integrating both sides: $\int \frac{1}{2(x - 1)} dx - \frac{1}{2} \int \frac{x}{x^2 + 1} dx - \frac{1}{2} \int \frac{1}{x^2 + 1} dx$.$= \frac{1}{2}\log |x - 1| - \frac{1}{4}\log (x^2 + 1) - \frac{1}{2}	an^{-1} x + c$.

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

Similar Questions

$\int {\frac{{{x^3} - 1}}{{{x^3} + x}}dx} = $

$\int \frac{x+1}{x^{2}+5 x+6} d x=$

$\int \frac{2 x^3-4 x^2-x-3}{x^2-2 x-3} d x=$

If $\int \frac{2 x^{2}+3}{\left(x^{2}-1\right)\left(x^{2}+4\right)} d x=a \log \left|\frac{x-1}{x+1}\right|+b \tan ^{-1}\left(\frac{x}{2}\right)+C$,then

Integrate the rational function: $\frac{1}{x^{4}-1}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module