What is the integral of (x+1)/(x^2+10x+34) ?

The integral of (x+1)/(x^2+10x+34) is 1/2 ln|x^2+10x+34|-4/3 arctan(1/3 (x+5))+C

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integral of (x+1)/(x^2+10x+34)

\int \frac{x + 1}{x ^{2} + 10 x + 34} d x

\frac{1}{2} ln x^{2} + 10 x + 34 - \frac{4}{3} arctan (\frac{1}{3} (x + 5)) + C

Solution steps

\int \frac{x + 1}{x ^{2} + 10 x + 34} d x

Complete the square

x^{2} + 10 x + 34 : (x + 5)^{2} + 9

= \int \frac{x + 1}{( x + 5 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{u - 4}{u ^{2} + 9} d u

= \int \frac{u - 4}{u ^{2} + 9} d u

Expand

\frac{u - 4}{u ^{2} + 9} : \frac{u}{u ^{2} + 9} - \frac{4}{u ^{2} + 9}

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} + 9} d u - \int \frac{4}{u ^{2} + 9} d u

\int \frac{u}{u ^{2} + 9} d u = \frac{1}{2} ln u^{2} + 9

\int \frac{4}{u ^{2} + 9} d u = \frac{4}{3} arctan (\frac{u}{3})

= \frac{1}{2} ln u^{2} + 9 - \frac{4}{3} arctan (\frac{u}{3})

Substitute back

u = x + 5

= \frac{1}{2} ln (x + 5)^{2} + 9 - \frac{4}{3} arctan (\frac{x + 5}{3})

Simplify

\frac{1}{2} ln (x + 5)^{2} + 9 - \frac{4}{3} arctan (\frac{x + 5}{3}) : \frac{1}{2} ln x^{2} + 10 x + 34 - \frac{4}{3} arctan (\frac{1}{3} (x + 5))

= \frac{1}{2} ln x^{2} + 10 x + 34 - \frac{4}{3} arctan (\frac{1}{3} (x + 5))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 10 x + 34 - \frac{4}{3} arctan (\frac{1}{3} (x + 5)) + C

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What is the integral of (x+1)/(x^2+10x+34) ?
The integral of (x+1)/(x^2+10x+34) is 1/2 ln|x^2+10x+34|-4/3 arctan(1/3 (x+5))+C