What is the integral of (x-4)/((x+5)^2) ?

The integral of (x-4)/((x+5)^2) is ln|x+5|+9/(x+5)+C

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integral of (x-4)/((x+5)^2)

\int \frac{x - 4}{( x + 5 ) ^{2}} d x

ln ∣ x + 5 ∣ + \frac{9}{x + 5} + C

Solution steps

\int \frac{x - 4}{( x + 5 ) ^{2}} d x

Apply u-substitution

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

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

Expand

\frac{u - 9}{u ^{2}} : \frac{1}{u} - \frac{9}{u ^{2}}

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= ln ∣ u ∣ - (- \frac{9}{u})

Substitute back

u = x + 5

= ln ∣ x + 5 ∣ - (- \frac{9}{x + 5})

Simplify

= ln ∣ x + 5 ∣ + \frac{9}{x + 5}

Add a constant to the solution

= ln ∣ x + 5 ∣ + \frac{9}{x + 5} + C

\int (2 x^{5} - x^{2})^{2} (x^{- \frac{1}{2}}) d x

\int \frac{1}{x ( 3 + x )} d x

\int (- 5 x + 1) d x

\int \frac{4 y}{x ^{3}} d x

What is the integral of (x-4)/((x+5)^2) ?
The integral of (x-4)/((x+5)^2) is ln|x+5|+9/(x+5)+C