int frac{x^5}{x^2+1} , dx = | vedclass

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

Question

(B) Let $I = \int \frac{x^5}{x^2+1} \, dx$. We can rewrite the integrand as: $\frac{x^5}{x^2+1} = \frac{x^3(x^2+1) - x^3}{x^2+1} = x^3 - \frac{x^3}{x^

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \frac{x^5}{x^2+1} \, dx$. We can rewrite the integrand as: $\frac{x^5}{x^2+1} = \frac{x^3(x^2+1) - x^3}{x^2+1} = x^3 - \frac{x^3}{x^2+1}$. Further,$\frac{x^3}{x^2+1} = \frac{x(x^2+1) - x}{x^2+1} = x - \frac{x}{x^2+1}$. So,$\frac{x^5}{x^2+1} = x^3 - x + \frac{x}{x^2+1}$. Now,integrating term by term: $I = \int (x^3 - x + \frac{x}{x^2+1}) \, dx = \frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{2} \int \frac{2x}{x^2+1} \, dx$. Using the substitution $u = x^2+1$,$du = 2x \, dx$,we get: $I = \frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{2} \log(x^2+1) + c$.

$\int \frac{x^5}{x^2+1} \, dx =$

Similar Questions

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

$\int a^x \, da = $

The value of $\int \sin 4x \cos 3x \, dx$ is

$\int \operatorname{cosec}(x-a) \cdot \operatorname{cosec} x \, dx = $

Find the following integral: $\int \frac{2-3 \sin x}{\cos ^{2} x} d x$

Vedclass Test Series

Exam Paper Generator

Online Exam Module