What is the integral of (x+6)/((x+1)^2) ?

The integral of (x+6)/((x+1)^2) is ln|x+1|-5/(x+1)+C

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integral of (x+6)/((x+1)^2)

\int \frac{x + 6}{( x + 1 ) ^{2}} d x

ln ∣ x + 1 ∣ - \frac{5}{x + 1} + C

Solution steps

\int \frac{x + 6}{( x + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{u + 5}{u ^{2}} d u

= \int \frac{u + 5}{u ^{2}} d u

Expand

\frac{u + 5}{u ^{2}} : \frac{1}{u} + \frac{5}{u ^{2}}

= \int \frac{1}{u} + \frac{5}{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{5}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{5}{u ^{2}} d u = - \frac{5}{u}

= ln ∣ u ∣ - \frac{5}{u}

Substitute back

u = x + 1

= ln ∣ x + 1 ∣ - \frac{5}{x + 1}

Add a constant to the solution

= ln ∣ x + 1 ∣ - \frac{5}{x + 1} + C

\frac{\partial ^{2}}{\partial r \partial θ} (r (θ) cos (θ))

x \to 0 lim (\frac{x ^{2} + 1}{x - 1})

(x + y + 1) d x - (y - x + 3) d y = 0

x \to - 1 lim (7 ∣ x + 1 ∣ - 4)

\int e^{2 x} d x

What is the integral of (x+6)/((x+1)^2) ?
The integral of (x+6)/((x+1)^2) is ln|x+1|-5/(x+1)+C