What is the integral of (2x+7)/(x^2+8x+16) ?

The integral of (2x+7)/(x^2+8x+16) is 2ln|x+4|+1/(x+4)+C

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integral of (2x+7)/(x^2+8x+16)

\int \frac{2 x + 7}{x ^{2} + 8 x + 16} d x

2 ln ∣ x + 4 ∣ + \frac{1}{x + 4} + C

Solution steps

\int \frac{2 x + 7}{x ^{2} + 8 x + 16} d x

Complete the square

x^{2} + 8 x + 16 : (x + 4)^{2}

= \int \frac{2 x + 7}{( x + 4 ) ^{2}} d x

Apply u-substitution

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

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

Expand

\frac{2 u - 1}{u ^{2}} : \frac{2}{u} - \frac{1}{u ^{2}}

= \int \frac{2}{u} - \frac{1}{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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

= 2 ln ∣ u ∣ - (- \frac{1}{u})

Substitute back

u = x + 4

= 2 ln ∣ x + 4 ∣ - (- \frac{1}{x + 4})

Simplify

= 2 ln ∣ x + 4 ∣ + \frac{1}{x + 4}

Add a constant to the solution

= 2 ln ∣ x + 4 ∣ + \frac{1}{x + 4} + C

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What is the integral of (2x+7)/(x^2+8x+16) ?
The integral of (2x+7)/(x^2+8x+16) is 2ln|x+4|+1/(x+4)+C