What is the integral of 8/(x^2(x^2+25)) ?

The integral of 8/(x^2(x^2+25)) is 8(-1/(25x)-1/125 arctan(x/5))+C

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integral of 8/(x^2(x^2+25))

\int \frac{8}{x ^{2} ( x ^{2} + 25 )} d x

8 (- \frac{1}{25 x} - \frac{1}{125} arctan (\frac{x}{5})) + C

Solution steps

\int \frac{8}{x ^{2} ( x ^{2} + 25 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int \frac{1}{x ^{2} ( x ^{2} + 25 )} d x

Take the partial fraction of

\frac{1}{x ^{2} ( x ^{2} + 25 )} : \frac{1}{25 x ^{2}} - \frac{1}{25 ( x ^{2} + 25 )}

= 8 \cdot \int \frac{1}{25 x ^{2}} - \frac{1}{25 ( x ^{2} + 25 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 (\int \frac{1}{25 x ^{2}} d x - \int \frac{1}{25 ( x ^{2} + 25 )} d x)

\int \frac{1}{25 x ^{2}} d x = - \frac{1}{25 x}

\int \frac{1}{25 ( x ^{2} + 25 )} d x = \frac{1}{125} arctan (\frac{x}{5})

= 8 (- \frac{1}{25 x} - \frac{1}{125} arctan (\frac{x}{5}))

Add a constant to the solution

= 8 (- \frac{1}{25 x} - \frac{1}{125} arctan (\frac{x}{5})) + C

f (x) = \frac{2}{ln ( x )}

x^{2} y y^{^{'}} = ey

\frac{d}{d x} ((\frac{1 - cos ( x )}{1 + cos ( x )})^{3})

x \to - 2 - lim (6 x + 5)

What is the integral of 8/(x^2(x^2+25)) ?
The integral of 8/(x^2(x^2+25)) is 8(-1/(25x)-1/125 arctan(x/5))+C