For any natural number m, evaluate ∫x3m+x2m+xm2x2m+3xm+61mdx, where x>0

I=∫x3m+x2m+xm2x2m+3xm+61mdxI=∫xx3m−1+x2m−1+xm−12x2m+3xm+61mdxPut 2x3m+3x2m+6xm=tm Now proceed.Aliter: I=∫ x3m+x2m+xm2x2m+3xm+61mdxLet xm = t ⇒ mxm – 1dx = dt∴ I=∫x⋅t3+t2+t2t2+3t+61mm⋅xm−1⋅xdt=∫x⋅t3+t2+t2t2+3t+61mm⋅xmdtI=∫tm⋅t3+t2+t2t2+3t+61mm⋅tdt=1m∫t3+t2+t2t3+3t2+6t1mtdt=1m∫t2+t+12t3+3t2+6t1mdtLet 2t3+3t2+6t=ym⇒6t2+t+1dt=mym−16dyI=1m∫m⋅ym−16⋅ydy=16∫ymdy=16ym−1(m+1)+C=162t3+3t2+6tm+1m+C =16(m+1)2x3m+3x2m+6xmm+1m+C

For any natural number m, evaluate ∫x3m+x2m+xm2x2m+3xm+61mdx, where x>0

I=∫x3m+x2m+xm2x2m+3xm+61mdxI=∫xx3m−1+x2m−1+xm−12x2m+3xm+61mdxPut 2x3m+3x2m+6xm=tm Now proceed.Aliter: I=∫ x3m+x2m+xm2x2m+3xm+61mdxLet xm = t ⇒ mxm – 1dx = dt∴ I=∫x⋅t3+t2+t2t2+3t+61mm⋅xm−1⋅xdt=∫x⋅t3+t2+t2t2+3t+61mm⋅xmdtI=∫tm⋅t3+t2+t2t2+3t+61mm⋅tdt=1m∫t3+t2+t2t3+3t2+6t1mtdt=1m∫t2+t+12t3+3t2+6t1mdtLet 2t3+3t2+6t=ym⇒6t2+t+1dt=mym−16dyI=1m∫m⋅ym−16⋅ydy=16∫ymdy=16ym−1(m+1)+C=162t3+3t2+6tm+1m+C =16(m+1)2x3m+3x2m+6xmm+1m+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

For any natural number $m$ , evaluate $\int (x^{3 m} + x^{2 m} + x^{m}) {(2 x^{2 m} + 3 x^{m} + 6)}^{\frac{1}{m}} d x$ , where $x > 0$

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

$\frac{1}{6 (m + 1)} {(2 x^{3 m} - 3 x^{2 m} - 6 x^{m})}^{\frac{m + 1}{m}} + C$

$\frac{1}{6} {(2 x^{3 m} + 3 x^{2 m} + 6 x^{m})}^{\frac{m + 1}{m}} + C$

$\frac{1}{6 (m - 1)} {(2 x^{3 m})}^{\frac{m - 1}{m}} + C$

$\frac{1}{6 (m + 1)} {(2 x^{3 m} + 3 x^{2 m} + 6 x^{m})}^{\frac{m + 1}{m}} + C$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$I = \int (x^{3 m} + x^{2 m} + x^{m}) {(2 x^{2 m} + 3 x^{m} + 6)}^{\frac{1}{m}} d x$

$I = \int x (x^{3 m - 1} + x^{2 m - 1} + x^{m - 1}) {(2 x^{2 m} + 3 x^{m} + 6)}^{\frac{1}{m}} d x$

Put $2 x^{3 m} + 3 x^{2 m} + 6 x^{m} = t^{m}$ Now proceed.

Aliter: $I = \int_{} (x^{3 m} + x^{2 m} + x^{m}) {(2 x^{2 m} + 3 x^{m} + 6)}^{\frac{1}{m}} dx$

Let $x^{m} = t \Rightarrow m x^{m - 1} d x = d t$

$\begin{array}{l} ∴ I = \int x \cdot \frac{(t^{3} + t^{2} + t) {(2 t^{2} + 3 t + 6)}^{\frac{1}{m}}}{m \cdot x^{m - 1} \cdot x} d t = \int \frac{x \cdot (t^{3} + t^{2} + t) {(2 t^{2} + 3 t + 6)}^{\frac{1}{m}}}{m \cdot x^{m}} d t \\ I = \int \frac{t^{m} \cdot (t^{3} + t^{2} + t) {(2 t^{2} + 3 t + 6)}^{\frac{1}{m}}}{m \cdot t} d t = \frac{1}{m} \int \frac{(t^{3} + t^{2} + t) {(2 t^{3} + 3 t^{2} + 6 t)}^{\frac{1}{m}}}{t} d t \end{array}$

$= \frac{1}{m} \int (t^{2} + t + 1) {(2 t^{3} + 3 t^{2} + 6 t)}^{\frac{1}{m}} d t$

Let $2 t^{3} + 3 t^{2} + 6 t = y^{m} \Rightarrow 6 (t^{2} + t + 1) d t = \frac{m y^{m - 1}}{6} d y$

$I = \frac{1}{m} \int \frac{m \cdot y^{m - 1}}{6} \cdot y d y = \frac{1}{6} \int y^{m} d y = \frac{1}{6} \frac{y^{m - 1}}{(m + 1)} + C = \frac{1}{6} {(2 t^{3} + 3 t^{2} + 6 t)}^{\frac{m + 1}{m}} + C = \frac{1}{6 (m + 1)} {(2 x^{3 m} + 3 x^{2 m} + 6 x^{m})}^{\frac{m + 1}{m}} + C$