∫0∞x3(1+x2)9/2 dx=

∫0∞ x3(1+x2)9/2 dx =∫0π/2 tan3θ(sec2θ)9/2sec2θ dθ put x=tanθ ⇒ dx=sec2 θ dθ x=0 ⇒θ=0 x=∞ ⇒ θ=π2 =∫0π/2 tan3θsec7 θ dθ =∫0π/2 sin3θ.co4θ dθ =(2) (3.1)7.5.3.1

∫0∞x3(1+x2)9/2 dx=

∫0∞ x3(1+x2)9/2 dx =∫0π/2 tan3θ(sec2θ)9/2sec2θ dθ put x=tanθ ⇒ dx=sec2 θ dθ x=0 ⇒θ=0 x=∞ ⇒ θ=π2 =∫0π/2 tan3θsec7 θ dθ =∫0π/2 sin3θ.co4θ dθ =(2) (3.1)7.5.3.1

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$\int_{0}^{\infty} \frac{x^{3}}{{(1 + x^{2})}^{9 / 2}} d x =$

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$\frac{3}{35}$

$\frac{1}{35}$

$\frac{2}{35}$

$\frac{4}{35}$

answer is C.

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Detailed Solution

$\int_{0}^{\infty} \frac{x^{3}}{{(1 + x^{2})}^{9 / 2}} dx = \int_{0}^{π / 2} \frac{\tan^{3} θ}{{(\sec^{2} θ)}^{9 / 2}} \sec^{2} θ dθ put x = tanθ \Rightarrow dx = \sec^{2} θ dθ x = 0 \Rightarrow θ = 0 x = \infty \Rightarrow θ = \frac{π}{2} = \int_{0}^{π / 2} \frac{\tan^{3} θ}{\sec^{7} θ} dθ = \int_{0}^{π / 2} \sin^{3} θ . {co}^{4} θ dθ = \frac{(2) (3.1)}{7.5.3.1}$