What is the integral of ((4x+16))/((x^2+6x+13)) ?

The integral of ((4x+16))/((x^2+6x+13)) is 2arctan(1/2 (x+3))+2ln|x^2+6x+13|+C

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integral of ((4x+16))/((x^2+6x+13))

integral

\frac{( 4 x + 16 )}{( x ^{2} + 6 x + 13 )}

2 arctan (\frac{1}{2} (x + 3)) + 2 ln x^{2} + 6 x + 13 + C

Solution steps

\int \frac{4 x + 16}{x ^{2} + 6 x + 13} d x

Complete the square

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

= \int \frac{4 x + 16}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

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

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

Apply Integration By Parts

: 2 arctan (\frac{u}{2}) (u + 1) - \int 2 arctan (\frac{u}{2}) d u

= 2 arctan (\frac{u}{2}) (u + 1) - \int 2 arctan (\frac{u}{2}) d u

\int 2 arctan (\frac{u}{2}) d u = 2 (u arctan (\frac{u}{2}) - ln u^{2} + 4)

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

Substitute back

u = x + 3

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

Simplify

2 arctan (\frac{x + 3}{2}) (x + 3 + 1) - 2 ((x + 3) arctan (\frac{x + 3}{2}) - ln (x + 3)^{2} + 4) : 2 arctan (\frac{1}{2} (x + 3)) + 2 ln x^{2} + 6 x + 13

= 2 arctan (\frac{1}{2} (x + 3)) + 2 ln x^{2} + 6 x + 13

Add a constant to the solution

= 2 arctan (\frac{1}{2} (x + 3)) + 2 ln x^{2} + 6 x + 13 + C

\frac{2 x + 1}{x}

(4 + x)^{- 1}

\frac{cos ^{5} ( x )}{sin ( x )}

y^{\frac{- 6}{7}}

\frac{( 5 x + 6 )}{( x ^{2} ( x + 3 ) )}

What is the integral of ((4x+16))/((x^2+6x+13)) ?
The integral of ((4x+16))/((x^2+6x+13)) is 2arctan(1/2 (x+3))+2ln|x^2+6x+13|+C