I=∫x4−xx201/4dx=∫x3−1x191/4dx=∫1x4x3−1x31/4dx=∫1x41−x−31+4dxPut 1−x−3=t. Then 3x−4dx=dt∴I=∫t1/4dt3=13t5/45/4+c=4151−x−35/4+c=415x3−1x35/4+c=415x3−15x151/4+c

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\int {[\frac{x^{4} - x}{x^{20}}]}^{1 / 4} d x =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{4}{15} {(\frac{{(x^{3} - 1)}^{5}}{x^{15}})}^{1 / 4} + e$

$\frac{4}{15} {(\frac{x^{4} + 1}{x^{4}})}^{1 / 4} + c$

$\frac{\sqrt{x^{4} + x^{2} + 1}}{x} + c$

$\frac{3}{4} (x^{4 / 3} + x^{1 / 3}) + c$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$I = \int {(\frac{x^{4} - x}{x^{20}})}^{1 / 4} d x = \int {(\frac{x^{3} - 1}{x^{19}})}^{1 / 4} d x = \int \frac{1}{x^{4}} {(\frac{x^{3} - 1}{x^{3}})}^{1 / 4} d x = \int \frac{1}{x^{4}} {(1 - x^{- 3})}^{1 + 4} d x$

Put $1 - x^{- 3} = t . Then 3 x^{- 4} d x = d t$

$∴ I = \int t^{1 / 4} \frac{d t}{3} = \frac{1}{3} \frac{t^{5 / 4}}{5 / 4} + c = \frac{4}{15} {(1 - x^{- 3})}^{5 / 4} + c = \frac{4}{15} {(\frac{x^{3} - 1}{x^{3}})}^{5 / 4} + c = \frac{4}{15} {(\frac{{(x^{3} - 1)}^{5}}{x^{15}})}^{1 / 4} + c$