What is the integral of 6/(x^2(x^2+25)) ?

The integral of 6/(x^2(x^2+25)) is 6(-1/(25x)-1/125 arctan(x/5))+C

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integral of 6/(x^2(x^2+25))

\int \frac{6}{x ^{2} ( x ^{2} + 25 )} d x

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

Solution steps

\int \frac{6}{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

= 6 \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 )}

= 6 \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

= 6 (\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})

= 6 (- \frac{1}{25 x} - \frac{1}{125} arctan (\frac{x}{5}))

Add a constant to the solution

= 6 (- \frac{1}{25 x} - \frac{1}{125} arctan (\frac{x}{5})) + C

x \to 0 lim (x^{2} + x + 1)

y = x^{6} f (x)

e^{x - y} y^{^{'}} = x - 3

\frac{\partial}{\partial z} (x^{2} + y^{2} z)

- 8 e^{π - t}

What is the integral of 6/(x^2(x^2+25)) ?
The integral of 6/(x^2(x^2+25)) is 6(-1/(25x)-1/125 arctan(x/5))+C