The value of the ∫(x2+x)(x−8+2x−9)110dx is

Let I=∫(x2+x)(x−8+2x−9)110dx =∫(x+1) (x2+2x)110dx . Put x2+2x=t then (x+1)dx=12dt=12∫(t)110dt

The value of the ∫(x2+x)(x−8+2x−9)110dx is

Let I=∫(x2+x)(x−8+2x−9)110dx =∫(x+1) (x2+2x)110dx . Put x2+2x=t then (x+1)dx=12dt=12∫(t)110dt

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The value of the $\int (x^{2} + x) {(x^{- 8} + 2 x^{- 9})}^{\frac{1}{10}} d x$ is

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$\frac{5}{11} {(x^{2} + 2 x)}^{\frac{11}{10}} + C$

$\frac{5}{6} {(x + 1)}^{\frac{11}{10}} + C$

$\frac{6}{7} {(x + 1)}^{\frac{11}{10}} + C$

$\frac{11}{5} {(x^{2} + 2 x)}^{\frac{10}{11}} + C$

answer is A.

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Detailed Solution

Let $I = \int (x^{2} + x) {(x^{- 8} + 2 x^{- 9})}^{\frac{1}{10}} d x$ $= \int (x + 1) {(x^{2} + 2 x)}^{\frac{1}{10}} d x$ . Put $x^{2} + 2 x = t$ then (x+1)dx= $\frac{1}{2}$ dt

$= \frac{1}{2} \int {(t)}^{\frac{1}{10}} d t$

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The value of the ∫(x2+x)(x−8+2x−9)110dx is

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The value of the $\int (x^{2} + x) {(x^{- 8} + 2 x^{- 9})}^{\frac{1}{10}} d x$ is