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

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

Question

(A) Let $I = \int \frac{x^3}{\sqrt{1+x^2}} dx = \int \frac{x^2 \cdot x}{\sqrt{1+x^2}} dx$. Substitute $t = \sqrt{1+x^2}$,so $t^2 = 1+x^2$ and $x^2 = t

Vedclass · Accepted Answer

$\frac{-2}{3}$

Vedclass · Answer

(A) Let $I = \int \frac{x^3}{\sqrt{1+x^2}} dx = \int \frac{x^2 \cdot x}{\sqrt{1+x^2}} dx$. Substitute $t = \sqrt{1+x^2}$,so $t^2 = 1+x^2$ and $x^2 = t^2 - 1$. Differentiating,$2t dt = 2x dx$,which implies $x dx = t dt$. Substituting these into the integral: $I = \int \frac{(t^2 - 1) t dt}{t} = \int (t^2 - 1) dt$. Integrating,we get $I = \frac{t^3}{3} - t + c$. Substituting back $t = \sqrt{1+x^2}$,we get $I = \frac{(1+x^2)^{\frac{3}{2}}}{3} - \sqrt{1+x^2} + c$. Comparing this with the given expression $a(1+x^2)^{\frac{3}{2}} + b \sqrt{1+x^2} + c$,we find $a = \frac{1}{3}$ and $b = -1$. Therefore,$a+b = \frac{1}{3} - 1 = \frac{-2}{3}$.

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

Similar Questions

The integral $\int \sec^{2/3} x \csc^{4/3} x \, dx$ is equal to: (Here $C$ is a constant of integration)

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

$\int \frac{d x}{(x+1) \sqrt{4 x+3}}$ is equal to

$\int \cos \sqrt{x} \, dx =$ (where $C$ is a constant of integration.)

$\int \frac{\ln |x|}{x\sqrt{1 + \ln |x|}} \, dx$ equals :

Vedclass Test Series

Exam Paper Generator

Online Exam Module