int {frac{{{x^5}dx}}{{sqrt {1 + {x^3}} }}} = | vedclass

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

Question

(D) Let $I = \int {\frac{{{x^5}dx}}{{\sqrt {1 + {x^3}} }}} $. We can write $x^5 dx$ as $x^3 \cdot x^2 dx$. Let $1 + x^3 = t^2$. Then $3x^2 dx = 2t dt$

Vedclass · Accepted Answer

$\frac{2}{9}\sqrt {1 + {x^3}} ({x^3} - 2) + C$

Vedclass · Answer

(D) Let $I = \int {\frac{{{x^5}dx}}{{\sqrt {1 + {x^3}} }}} $. We can write $x^5 dx$ as $x^3 \cdot x^2 dx$. Let $1 + x^3 = t^2$. Then $3x^2 dx = 2t dt$,which implies $x^2 dx = \frac{2}{3}t dt$. Also,$x^3 = t^2 - 1$. Substituting these into the integral: $I = \int {\frac{{(t^2 - 1)}}{t} \cdot \frac{2}{3}t dt} = \frac{2}{3} \int {(t^2 - 1) dt} $. Integrating with respect to $t$: $I = \frac{2}{3} \left( \frac{t^3}{3} - t \right) + C = \frac{2}{9} t(t^2 - 3) + C$. Since $t = \sqrt{1 + x^3}$ and $t^2 = 1 + x^3$,we have: $I = \frac{2}{9} \sqrt{1 + x^3} (1 + x^3 - 3) + C = \frac{2}{9} \sqrt{1 + x^3} (x^3 - 2) + C$.

$\int {\frac{{{x^5}dx}}{{\sqrt {1 + {x^3}} }}} = $

Similar Questions

$\int {\sqrt {{e^x} - 1} } dx = $

If $\int \frac{\cos x}{\sqrt{4 \sin ^2 x+4 \sin x+5}} d x=\frac{1}{2} \sinh ^{-1}(f(x))+C$, then find $2 f(x)$.

$\int \left( \frac{4 \tan^4 x + 3 \tan^2 x - 1}{\tan^2 x + 4} \right) dx =$

$\int \frac{\sin (\tan ^{-1} x)}{1+x^2} d x=$ . . . . . . $+C$.

$\int \frac{\sqrt{\cot x}}{\sin 2x} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module