Evaluate -x7 d x1-x25where C is constant of integration

Let I=∫x^7 d x(1 x^2)^5=∫x^7 d xx^10(x^ 2 1)^5 ⇒ I=∫x^ 3 d x(x^ 2 1)^5 Put : x^ 2 1=t nbsp; ⇒ 2x^ 3dx=dt ⇒ I= 12∫d t nbsp;t^5 ⇒ I= 12∫ t^ 5 d t=12·t^ 44+C ...

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Byju's Answer

Standard XII

Mathematics

Binomial Coefficients

Evaluate ∫x7 ...

Question

Evaluate $\int \frac{x^{7} d x}{{(1 - x^{2})}^{5}}$
(where $C$ is constant of integration)

- \frac{x^{8}}{8 {(1 - x^{2})}^{4}} + C

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\frac{x^{8}}{8 {(1 - x^{2})}^{4}} + C

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\frac{x^{4}}{8 {(1 - x^{2})}^{4}} + C

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- \frac{x^{4}}{8 {(1 - x^{2})}^{4}} + C

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Solution

The correct option is B $\frac{x^{8}}{8 {(1 - x^{2})}^{4}} + C$
Let $I = \int \frac{x^{7} d x}{{(1 - x^{2})}^{5}} = \int \frac{x^{7} d x}{x^{10} {(x^{- 2} - 1)}^{5}}$
$\Rightarrow I = \int \frac{x^{- 3} d x}{{(x^{- 2} - 1)}^{5}}$
Put : $x^{- 2} - 1 = t$
$\Rightarrow - 2 x^{- 3} d x = d t$
$\Rightarrow I = - \frac{1}{2} \int \frac{d t}{t^{5}}$

$\Rightarrow I = - \frac{1}{2} \int t^{- 5} d t = \frac{1}{2} \cdot \frac{t^{- 4}}{4} + C$

$\Rightarrow I = \frac{1}{8} \frac{1}{{(x^{- 2} - 1)}^{4}} + C$

$∴ I = \frac{x^{8}}{8 {(1 - x^{2})}^{4}} + C$

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Binomial Coefficients

Standard XII Mathematics

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Evaluate ∫x7dx(1−x2)5 (where C is constant of integration)

The correct option is B x88(1−x2)4+CLet I=∫x7dx(1−x2)5=∫x7dxx10(x−2−1)5 ⇒I=∫x−3dx(x−2−1)5 Put : x−2−1=t ⇒−2x−3dx=dt ⇒I=−12∫dtt5 ⇒I=−12∫t−5dt=12⋅t−44+C ⇒I=181(x−2−1)4+C ∴I=x88(1−x2)4+C

Evaluate $\int \frac{x^{7} d x}{{(1 - x^{2})}^{5}}$
(where $C$ is constant of integration)