Let f(x)=x(1+xn)1/n for n>2 and g(x)=fofo.....of(x)⏟f occurs n times. then ∫xn−2g(x)dx equal to

∫xn−2g(x)dx=∫xn−1dx(1+nxn)1/n=1n2∫n2xn−1dx(1+nxn)1/n=1n2∫ddx(1+nxn)(1+nxn)1/n

Let f(x)=x(1+xn)1/n for n>2 and g(x)=fofo.....of(x)⏟f occurs n times. then ∫xn−2g(x)dx equal to

∫xn−2g(x)dx=∫xn−1dx(1+nxn)1/n=1n2∫n2xn−1dx(1+nxn)1/n=1n2∫ddx(1+nxn)(1+nxn)1/n

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

Let $f (x) = \frac{x}{{(1 + x^{n})}^{1 / n}}$ for $n > 2$ and $g (x) = \underset{f o c c u r s n t i m e s}{\underset{⏟}{f o f o ..... o f (x)}} .$ then $\int x^{n - 2} g (x) d x$ equal to

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

$\frac{1}{(n - 1)} {(1 + n x^{n})}^{1 - \frac{1}{n}} + K$

$\frac{1}{n (n - 1)} {(1 + n x^{n})}^{1 - \frac{1}{n}} + K$

$\frac{1}{(n + 1)} {(1 + n x^{n})}^{1 + \frac{1}{n}} + K$

$\frac{1}{n (n - 1)} {(1 + n x^{n})}^{1 + \frac{1}{n}} + K$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \int x^{n - 2} g (x) d x = \int \frac{x^{n - 1} d x}{{(1 + n x^{n})}^{1 / n}} \\ = \frac{1}{n^{2}} \int \frac{n^{2} x^{n - 1} d x}{{(1 + n x^{n})}^{1 / n}} = \frac{1}{n^{2}} \int \frac{\frac{d}{d x} (1 + n x^{n})}{{(1 + n x^{n})}^{1 / n}} \end{array}$