The value of the integral ∫x2+xx−8+2x−91/10dx is

I=∫x2+xx−8+2x−91/10dx=∫(x+1)x2+2x1/10dx Put x2+2x=t⇒(x+1)dx=dt2 Now, I=∫t110⋅dt2=12×1011t1110=511t1110+c∴ I=511x2+2x1110+c

The value of the integral ∫x2+xx−8+2x−91/10dx is

I=∫x2+xx−8+2x−91/10dx=∫(x+1)x2+2x1/10dx Put x2+2x=t⇒(x+1)dx=dt2 Now, I=∫t110⋅dt2=12×1011t1110=511t1110+c∴ I=511x2+2x1110+c

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

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$None of these$

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

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int (x^{2} + x) {(x^{- 8} + 2 x^{- 9})}^{1 / 10} d x \\ = \int (x + 1) {(x^{2} + 2 x)}^{1 / 10} d x \\ Put x^{2} + 2 x = t \Rightarrow (x + 1) d x = \frac{d t}{2} \\ Now, I = \int t^{\frac{1}{10}} \cdot \frac{d t}{2} = \frac{1}{2} \times \frac{10}{11} t^{\frac{11}{10}} = \frac{5}{11} t^{\frac{11}{10}} + c \\ ∴ I = \frac{5}{11} {(x^{2} + 2 x)}^{\frac{11}{10}} + c \end{array}$