What is the integral of x/(x^2+10x+50) ?

The integral of x/(x^2+10x+50) is 1/2 ln|x^2+10x+50|-arctan(1/5 (x+5))+C

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integral of x/(x^2+10x+50)

\int \frac{x}{x ^{2} + 10 x + 50} d x

\frac{1}{2} ln x^{2} + 10 x + 50 - arctan (\frac{1}{5} (x + 5)) + C

Solution steps

\int \frac{x}{x ^{2} + 10 x + 50} d x

Complete the square

x^{2} + 10 x + 50 : (x + 5)^{2} + 25

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

Apply u-substitution

: \int \frac{u - 5}{u ^{2} + 25} d u

= \int \frac{u - 5}{u ^{2} + 25} d u

Expand

\frac{u - 5}{u ^{2} + 25} : \frac{u}{u ^{2} + 25} - \frac{5}{u ^{2} + 25}

Apply the Sum Rule:

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

= \int \frac{u}{u ^{2} + 25} d u - \int \frac{5}{u ^{2} + 25} d u

\int \frac{u}{u ^{2} + 25} d u = \frac{1}{2} ln u^{2} + 25

\int \frac{5}{u ^{2} + 25} d u = arctan (\frac{u}{5})

= \frac{1}{2} ln u^{2} + 25 - arctan (\frac{u}{5})

Substitute back

u = x + 5

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

Simplify

\frac{1}{2} ln (x + 5)^{2} + 25 - arctan (\frac{x + 5}{5}) : \frac{1}{2} ln x^{2} + 10 x + 50 - arctan (\frac{1}{5} (x + 5))

= \frac{1}{2} ln x^{2} + 10 x + 50 - arctan (\frac{1}{5} (x + 5))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 10 x + 50 - arctan (\frac{1}{5} (x + 5)) + C

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What is the integral of x/(x^2+10x+50) ?
The integral of x/(x^2+10x+50) is 1/2 ln|x^2+10x+50|-arctan(1/5 (x+5))+C