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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln ∣ x + 4 ∣ + \frac{9}{x + 4} + C

Solution steps

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

= ln ∣ x + 4 ∣ - (- \frac{9}{x + 4})

Simplify

= ln ∣ x + 4 ∣ + \frac{9}{x + 4}

Add a constant to the solution

= ln ∣ x + 4 ∣ + \frac{9}{x + 4} + C

\int \frac{x ^{2} - 1}{x ^{2} - 16} d x

\int x^{5} \cdot e^{2 x^{3}} d x

\int \frac{- 0.33 x}{0.02 x ^{2} + 10} d x

\int 6 sin (x) cos^{4} (x) d x

\int \frac{z ^{2}}{e ^{- z^{3}}} d z

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