int {{e^{3log x}}{{({x^4} + 1)}^{ - 1}},dx} = | vedclass

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

Question

(B) Given integral: $I = \int {{e^{3\log x}}{{({x^4} + 1)}^{ - 1}}dx}$Using the property $n\log x = \log(x^n)$ and $e^{\log f(x)} = f(x)$,we have:$e^{

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given integral: $I = \int {{e^{3\log x}}{{({x^4} + 1)}^{ - 1}}dx}$Using the property $n\log x = \log(x^n)$ and $e^{\log f(x)} = f(x)$,we have:$e^{3\log x} = e^{\log(x^3)} = x^3$So,the integral becomes:$I = \int \frac{x^3}{x^4 + 1} dx$Let $u = x^4 + 1$. Then $du = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} du$.Substituting these into the integral:$I = \int \frac{1}{u} \cdot \frac{1}{4} du = \frac{1}{4} \int \frac{1}{u} du$$I = \frac{1}{4} \log|u| + c$Substituting back $u = x^4 + 1$:$I = \frac{1}{4} \log(x^4 + 1) + c$

$\int {{e^{3\log x}}{{({x^4} + 1)}^{ - 1}}\,dx} = $

Similar Questions

$\int \frac{d x}{(x+a)^{\frac{9}{7}}(x-b)^{\frac{5}{7}}} = ?$

$\int \sqrt{e^x-4} \, dx$ equals

$\int \frac{dx}{x^2 \sqrt{4+x^2}}$ is equal to

Evaluate the integral $\int \frac{\sqrt{\cot x}}{\sin x \cos x} d x = -f(x) + c$. Find $f(x)$.

$\int \sqrt{\operatorname{cosec} x - \sin x} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module