How do you find the antiderivative of $\frac{1}{{(x^{2} + 25)}^{2}} d x$ $1/((x^2+25)^2) dx$ ?

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How do you find the antiderivative of $\frac{1}{{(x^{2} + 25)}^{2}} d x$ $1/((x^2+25)^2) dx$ ?

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Answer 1 · 2015-09-13T03:11:38

$\frac{d y}{d x} = \frac{- 4 x}{{(x^{2} + 25)}^{- 3}}$ $dy/dx =( -4x)/(x^2+25)^(-3)$

Explanation:

In order to have a simple solution, the above expression can be written as

$y = {(x^{2} + 25)}^{- 2}$ $y = (x^2+25)^(-2)$

Now use function of a function rule

$\frac{d y}{d x} = - 2 {(x^{2} + 25)}^{- 3} (2 x)$ $dy/dx = -2(x^2+25)^(-3)(2x)$
$\frac{d y}{d x} = - 4 x {(x^{2} + 25)}^{- 3}$ $dy/dx = -4x(x^2+25)^(-3)$
$\frac{d y}{d x} = \frac{- 4 x}{{(x^{2} + 25)}^{- 3}}$ $dy/dx =( -4x)/(x^2+25)^(-3)$

You can use the quotient rule also

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How do you find the antiderivative of $\frac{1}{{(x^{2} + 25)}^{2}} d x$ $1/((x^2+25)^2) dx$ ?

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Explanation:

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How do you find the antiderivative of $\frac{1}{{(x^{2} + 25)}^{2}} d x$ $1/((x^2+25)^2) dx$ ?