What is the integral of (3x+7)/(x^2+6x+25) ?

The integral of (3x+7)/(x^2+6x+25) is 3/2 ln|x^2+6x+25|-1/2 arctan(1/4 (x+3))+C

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integral of (3x+7)/(x^2+6x+25)

\int \frac{3 x + 7}{x ^{2} + 6 x + 25} d x

\frac{3}{2} ln x^{2} + 6 x + 25 - \frac{1}{2} arctan (\frac{1}{4} (x + 3)) + C

Solution steps

\int \frac{3 x + 7}{x ^{2} + 6 x + 25} d x

Complete the square

x^{2} + 6 x + 25 : (x + 3)^{2} + 16

= \int \frac{3 x + 7}{( x + 3 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{3 u - 2}{u ^{2} + 16} d u

= \int \frac{3 u - 2}{u ^{2} + 16} d u

Expand

\frac{3 u - 2}{u ^{2} + 16} : \frac{3 u}{u ^{2} + 16} - \frac{2}{u ^{2} + 16}

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{3 u}{u ^{2} + 16} d u - \int \frac{2}{u ^{2} + 16} d u

\int \frac{3 u}{u ^{2} + 16} d u = \frac{3}{2} ln u^{2} + 16

\int \frac{2}{u ^{2} + 16} d u = \frac{1}{2} arctan (\frac{u}{4})

= \frac{3}{2} ln u^{2} + 16 - \frac{1}{2} arctan (\frac{u}{4})

Substitute back

u = x + 3

= \frac{3}{2} ln (x + 3)^{2} + 16 - \frac{1}{2} arctan (\frac{x + 3}{4})

Simplify

\frac{3}{2} ln (x + 3)^{2} + 16 - \frac{1}{2} arctan (\frac{x + 3}{4}) : \frac{3}{2} ln x^{2} + 6 x + 25 - \frac{1}{2} arctan (\frac{1}{4} (x + 3))

= \frac{3}{2} ln x^{2} + 6 x + 25 - \frac{1}{2} arctan (\frac{1}{4} (x + 3))

Add a constant to the solution

= \frac{3}{2} ln x^{2} + 6 x + 25 - \frac{1}{2} arctan (\frac{1}{4} (x + 3)) + C

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What is the integral of (3x+7)/(x^2+6x+25) ?
The integral of (3x+7)/(x^2+6x+25) is 3/2 ln|x^2+6x+25|-1/2 arctan(1/4 (x+3))+C