What is the integral of 5/(x(x^2+5)^2) ?

The integral of 5/(x(x^2+5)^2) is 1/5 ln(x)-1/10 ln|x^2+5|+1/(2(x^2+5))+C

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integral of 5/(x(x^2+5)^2)

\int \frac{5}{x ( x ^{2} + 5 ) ^{2}} d x

\frac{1}{5} ln (x) - \frac{1}{10} ln x^{2} + 5 + \frac{1}{2 ( x ^{2} + 5 )} + C

Solution steps

\int \frac{5}{x ( x ^{2} + 5 ) ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ( x ^{2} + 5 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{2 ( u - 5 ) u ^{2}} d u

= 5 \cdot \int \frac{1}{2 ( u - 5 ) u ^{2}} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{( u - 5 ) u ^{2}} d u

Take the partial fraction of

\frac{1}{( u - 5 ) u ^{2}} : \frac{1}{25 ( u - 5 )} - \frac{1}{25 u} - \frac{1}{5 u ^{2}}

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{25 ( u - 5 )} - \frac{1}{25 u} - \frac{1}{5 u ^{2}} d u

Apply the Sum Rule:

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

= 5 \cdot \frac{1}{2} (\int \frac{1}{25 ( u - 5 )} d u - \int \frac{1}{25 u} d u - \int \frac{1}{5 u ^{2}} d u)

\int \frac{1}{25 ( u - 5 )} d u = \frac{1}{25} ln ∣ u - 5 ∣

\int \frac{1}{25 u} d u = \frac{1}{25} ln ∣ u ∣

\int \frac{1}{5 u ^{2}} d u = - \frac{1}{5 u}

= 5 \cdot \frac{1}{2} (\frac{1}{25} ln ∣ u - 5 ∣ - \frac{1}{25} ln ∣ u ∣ - (- \frac{1}{5 u}))

Substitute back

u = x^{2} + 5

= 5 \cdot \frac{1}{2} (\frac{1}{25} ln x^{2} + 5 - 5 - \frac{1}{25} ln x^{2} + 5 - (- \frac{1}{5 ( x ^{2} + 5 )}))

Simplify

5 \cdot \frac{1}{2} (\frac{1}{25} ln x^{2} + 5 - 5 - \frac{1}{25} ln x^{2} + 5 - (- \frac{1}{5 ( x ^{2} + 5 )})) : \frac{1}{5} ln (x) - \frac{1}{10} ln x^{2} + 5 + \frac{1}{2 ( x ^{2} + 5 )}

= \frac{1}{5} ln (x) - \frac{1}{10} ln x^{2} + 5 + \frac{1}{2 ( x ^{2} + 5 )}

Add a constant to the solution

= \frac{1}{5} ln (x) - \frac{1}{10} ln x^{2} + 5 + \frac{1}{2 ( x ^{2} + 5 )} + C

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What is the integral of 5/(x(x^2+5)^2) ?
The integral of 5/(x(x^2+5)^2) is 1/5 ln(x)-1/10 ln|x^2+5|+1/(2(x^2+5))+C