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

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

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

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

5 ln ∣ x + 1 ∣ + x + C

Solution steps

\int \frac{x + 6}{x + 1} d x

\frac{x + 6}{x + 1} = \frac{5}{x + 1} + 1

= \int \frac{5}{x + 1} + 1 d x

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{5}{x + 1} d x + \int 1 d x

\int \frac{5}{x + 1} d x = 5 ln ∣ x + 1 ∣

\int 1 d x = x

= 5 ln ∣ x + 1 ∣ + x

Add a constant to the solution

= 5 ln ∣ x + 1 ∣ + x + C

\int_{2}^{3} \frac{ln ( 2 x ^{5} )}{x ^{2}} d x

\frac{\partial}{\partial x} (x^{a} y^{1 - a})

\frac{\partial}{\partial x} (sin (x + 2 y))

\int_{- 1}^{1} x (x^{2} + 3)^{3} d x

x \to - 2 - lim (\frac{3 x - 6}{x + 2})

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