If int sqrt[3]{x}left{1+sqrt[3]{x^4}right}^{1 | vedclass

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

Question

(B) Let $I = \int \sqrt[3]{x}\left(1+\sqrt[3]{x^4}\right)^{\frac{1}{7}} d x = \int x^{\frac{1}{3}}\left(1+x^{\frac{4}{3}}\right)^{\frac{1}{7}} d x$. L

Vedclass · Accepted Answer

$3 / 4$

Vedclass · Answer

(B) Let $I = \int \sqrt[3]{x}\left(1+\sqrt[3]{x^4}ight)^{\frac{1}{7}} d x = \int x^{\frac{1}{3}}\left(1+x^{\frac{4}{3}}ight)^{\frac{1}{7}} d x$. Let $1+x^{\frac{4}{3}} = t$. Then,$\frac{4}{3} x^{\frac{1}{3}} d x = d t$,which implies $x^{\frac{1}{3}} d x = \frac{3}{4} d t$. Substituting these into the integral,we get: $I = \int t^{\frac{1}{7}} \cdot \frac{3}{4} d t = \frac{3}{4} \int t^{\frac{1}{7}} d t$. Integrating,we have: $I = \frac{3}{4} \cdot \frac{t^{\frac{1}{7}+1}}{\frac{1}{7}+1} + c = \frac{3}{4} \cdot \frac{t^{\frac{8}{7}}}{\frac{8}{7}} + c = \frac{3}{4} \cdot \frac{7}{8} t^{\frac{8}{7}} + c = \frac{21}{32} \left(1+x^{\frac{4}{3}}ight)^{\frac{8}{7}} + c$. Comparing this with $A\left(1+\sqrt[3]{x^4}ight)^B+c$,we get $A = \frac{21}{32}$ and $B = \frac{8}{7}$. Therefore,$A B = \frac{21}{32} 	imes \frac{8}{7} = \frac{3}{4}$.

If $\int \sqrt[3]{x}\left\{1+\sqrt[3]{x^4}\right\}^{1 / 7} d x=A\left(1+\sqrt[3]{x^4}\right)^B+c$,then the value of $A B$ is equal to

Similar Questions

$\int \sqrt{\frac{x}{a^3 - x^3}} \, dx = $

$\int \cos ^3 x e^{\log (\sin x)^2} d x=$

$\int \sin \sqrt{x} \,d x=\ldots+C$ (where $C$ is a constant of integration.)

The value of the integral $\int \frac{\sin(\ln(2 + 2x))}{x + 1} dx$ is

$\int \sin^2 x \cos x \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module