Test Index

AP EAPCET 24-Aug-2021 Shift 1 Paper

Section: Mathematics

© examsnet.com

Question : 76 of 160

Marks: +1, -0

If

\int \sqrt[3]{x} \left(1+\sqrt[3]{x^{4}}\right)^{\frac{1}{7}} \, dx = A\left(1+\sqrt[3]{x^{4}}\right)^{B}+C

, then value of

AB=

_____

$\frac{3}{2}$
$\frac{3}{4}$
$\frac{3}{32}$
$\frac{4}{3}$

Solution:

\int \sqrt[3]{x} \left(1+\sqrt[3]{x^{4}}\right)^{\frac{1}{7}} \, dx = A\left(1+\sqrt[3]{x^{4}}\right)^{B}+C

Let us assume

I = \int \sqrt[3]{x} \left(1+\sqrt[3]{x^{4}}\right)^{\frac{1}{7}} \, dx = \int x^{\frac{1}{3}} \left(1+x^{\frac{4}{3}}\right)^{\frac{1}{7}} \, dx

Let

1+x^{\frac{4}{3}}=t

\Rightarrow \frac{4}{3} x^{\frac{4}{3}-1} dx = dt

\Rightarrow \frac{4}{3} x^{\frac{1}{3}} dx = dt

\Rightarrow x^{\frac{1}{3}} dx = \frac{3}{4} dt

∴

I = \int (t)^{\frac{1}{7}} \cdot \frac{3}{4} \, dt = \frac{3}{4} \int (t)^{\frac{1}{7}} \, dt

= \frac{3}{4} \left[ \frac{(t)^{\frac{1}{7}+1}}{\frac{1}{7}+1} \right] + c = \frac{3}{4} \frac{(1+x^{4/3})^{\frac{8}{7}}}{\frac{8}{7}} + c

= \frac{21}{32} (1+\sqrt[3]{x^{4}})^{\frac{8}{7}} + c

∴

A = \frac{21}{32}

and

B = \frac{8}{7}

∴

AB = \frac{21}{32} \times \frac{8}{7} = \frac{3}{4}

© examsnet.com

Go to Question:

Prev Question

Next Question