Given-0-dx-a-2-x-2-pi-2a-find-0-dx-a-2-x-2-3-

Question Number 98826 by bramlex last updated on 16/Jun/20

G i v e n \underset{0}{\int^{\infty}} \frac{d x}{a^{2} + x^{2}} = \frac{π}{2 a}

f i n d \underset{0}{\int^{\infty}} \frac{d x}{{(a^{2} + x^{2})}^{3}} ?

Commented by bemath last updated on 16/Jun/20

\frac{d}{da} [\underset{0}{\int^{\infty}} \frac{dx}{a^{2} + x^{2}}] = \frac{d}{da} [\frac{π}{2 a}]

- 2 a \underset{0}{\int^{\infty}} \frac{dx}{{(a^{2} + x^{2})}^{2}} = \frac{- π}{{2 a}^{2}}

\underset{0}{\int^{\infty}} \frac{dx}{{(a^{2} + x^{2})}^{2}} = \frac{π}{{4 a}^{3}}

\frac{d}{da} [\underset{0}{\int^{\infty}} \frac{dx}{{(a^{2} + x^{2})}^{2}}] = \frac{d}{da} [\frac{π}{{4 a}^{3}}]

- 4 a \underset{0}{\int^{\infty}} \frac{dx}{{(a^{2} + x^{2})}^{3}} = \frac{- 3 π}{{4 a}^{4}}

∴ \underset{0}{\int^{\infty}} \frac{dx}{{(a^{2} + x^{2})}^{3}} = \frac{3 π}{{16 a}^{5}}

Commented by bramlex last updated on 16/Jun/20

g r e a t t t

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