If int frac{x^3 , dx}{sqrt{1+x^2}} = a(1+x^2) | vedclass

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

Question

(B) Let $I = \int \frac{x^3 \, dx}{\sqrt{1+x^2}}$.Put $1+x^2 = t^2$,then $2x \, dx = 2t \, dt$,which implies $x \, dx = t \, dt$.Also,$x^2 = t^2 - 1$.

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) Let $I = \int \frac{x^3 \, dx}{\sqrt{1+x^2}}$.Put $1+x^2 = t^2$,then $2x \, dx = 2t \, dt$,which implies $x \, dx = t \, dt$.Also,$x^2 = t^2 - 1$.Substituting these into the integral:$I = \int \frac{(t^2-1) \cdot t \, dt}{t} = \int (t^2 - 1) \, dt$.Integrating with respect to $t$:$I = \frac{t^3}{3} - t + c$.Substituting $t = \sqrt{1+x^2}$ back:$I = \frac{(1+x^2)^{3/2}}{3} - \sqrt{1+x^2} + c = \frac{1}{3}(1+x^2)\sqrt{1+x^2} - 1\sqrt{1+x^2} + c$.Comparing this with the given form $a(1+x^2)\sqrt{1+x^2} + b\sqrt{1+x^2} + c$,we get $a = \frac{1}{3}$ and $b = -1$.Therefore,$3ab = 3 	imes \frac{1}{3} 	imes (-1) = -1$.

If $\int \frac{x^3 \, dx}{\sqrt{1+x^2}} = a(1+x^2) \sqrt{1+x^2} + b \sqrt{1+x^2} + c$ (where $c$ is a constant of integration),then the value of $3ab$ is

Similar Questions

$\int \cos (\log x) d x=F(x)+C,$ where $C$ is an arbitrary constant. Here, $F(x)$ is equal to

$\int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \,d x=$ (where $C$ is a constant of integration.)

$\int \frac{1+x \cos x}{x\left[1-x^2\left(e^{\sin x}\right)^2\right]} d x=$

$\int {\frac{{\sqrt {{x^2} + 1} [\log ({x^2} + 1) - 2\log x]}}{{{x^4}}}} dx$ is equal to

$\int (x^{21} + x^6 + x^3)(2x^{18} + 7x^3 + 14)^{1/3} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module