What is the integral of (2x+32)/(x^2+8x+7) ?

The integral of (2x+32)/(x^2+8x+7) is -3ln|x+7|+5ln|x+1|+C

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integral of (2x+32)/(x^2+8x+7)

\int \frac{2 x + 32}{x ^{2} + 8 x + 7} d x

- 3 ln ∣ x + 7 ∣ + 5 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{2 x + 32}{x ^{2} + 8 x + 7} d x

Complete the square

x^{2} + 8 x + 7 : (x + 4)^{2} - 9

= \int \frac{2 x + 32}{( x + 4 ) ^{2} - 9} d x

Apply u-substitution

: \int \frac{2 u + 24}{u ^{2} - 9} d u

= \int \frac{2 u + 24}{u ^{2} - 9} d u

Take the partial fraction of

\frac{2 u + 24}{u ^{2} - 9} : - \frac{3}{u + 3} + \frac{5}{u - 3}

= \int - \frac{3}{u + 3} + \frac{5}{u - 3} 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{3}{u + 3} d u + \int \frac{5}{u - 3} d u

\int \frac{3}{u + 3} d u = 3 ln ∣ u + 3 ∣

\int \frac{5}{u - 3} d u = 5 ln ∣ u - 3 ∣

= - 3 ln ∣ u + 3 ∣ + 5 ln ∣ u - 3 ∣

Substitute back

u = x + 4

= - 3 ln ∣ x + 4 + 3 ∣ + 5 ln ∣ x + 4 - 3 ∣

Simplify

= - 3 ln ∣ x + 7 ∣ + 5 ln ∣ x + 1 ∣

Add a constant to the solution

= - 3 ln ∣ x + 7 ∣ + 5 ln ∣ x + 1 ∣ + C

y = \frac{1}{2} x^{3}

\frac{d}{d x} (\frac{1}{( x + 7 ) ^{2}})

f (x) = x \cdot tan (x^{2})

\frac{\partial}{\partial y} (3 x^{2} y + e^{y})

y = 8 x + 6 x^{\frac{3}{4}}

What is the integral of (2x+32)/(x^2+8x+7) ?
The integral of (2x+32)/(x^2+8x+7) is -3ln|x+7|+5ln|x+1|+C