Consider f(x) = frac{x^2}{1 + x^3} and g(t) = | vedclass

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

Question

(B) Given $f(x) = \frac{x^2}{1 + x^3}$.We need to find $g(x) = \int f(x) \, dx = \int \frac{x^2}{1 + x^3} \, dx$.Let $u = 1 + x^3$. Then $du = 3x^2 \,

Vedclass · Accepted Answer

$\frac{1}{3} \ln\left( \frac{1 + x^3}{2} \right)$

Vedclass · Answer

(B) Given $f(x) = \frac{x^2}{1 + x^3}$.We need to find $g(x) = \int f(x) \, dx = \int \frac{x^2}{1 + x^3} \, dx$.Let $u = 1 + x^3$. Then $du = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{du}{3}$.Substituting these into the integral:$g(x) = \int \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \ln|u| + C = \frac{1}{3} \ln(1 + x^3) + C$.Given the condition $g(1) = 0$:$0 = \frac{1}{3} \ln(1 + 1^3) + C \implies 0 = \frac{1}{3} \ln(2) + C \implies C = -\frac{1}{3} \ln(2)$.Substituting $C$ back into the expression for $g(x)$:$g(x) = \frac{1}{3} \ln(1 + x^3) - \frac{1}{3} \ln(2) = \frac{1}{3} \ln\left( \frac{1 + x^3}{2} \right)$.

Consider $f(x) = \frac{x^2}{1 + x^3}$ and $g(t) = \int f(t) \, dt$. If $g(1) = 0$,then $g(x)$ equals:

Similar Questions

$\int \frac{x}{1 + x^4} \, dx = $

If $\int_1^4 x \sqrt{x^2-1} \, dx = \alpha(k)^\beta$,then $\alpha \beta$ is equal to

$\int(x+2) \sqrt{x+3} \, dx =$

Let $n \ge 2$ be a natural number and $0 < \theta < \frac{\pi}{2}$. Then $\int \frac{(\sin^n \theta - \sin \theta)^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} d\theta$ is equal to

$\int \frac{3^x \, dx}{\sqrt{9^x-1}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module