What is the integral of 8/(x+1)+x ?

The integral of 8/(x+1)+x is 8ln|x+1|+(x^2)/2+C

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integral of 8/(x+1)+x

\int \frac{8}{x + 1} + x d x

8 ln ∣ x + 1 ∣ + \frac{x ^{2}}{2} + C

Solution steps

\int \frac{8}{x + 1} + x 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{8}{x + 1} d x + \int x d x

\int \frac{8}{x + 1} d x = 8 ln ∣ x + 1 ∣

\int x d x = \frac{x ^{2}}{2}

= 8 ln ∣ x + 1 ∣ + \frac{x ^{2}}{2}

Add a constant to the solution

= 8 ln ∣ x + 1 ∣ + \frac{x ^{2}}{2} + C

\int \frac{5 x - 12}{x ^{2} - 4 x} d x

\frac{d}{d x} ((3 x + 7)^{2})

\int_{- 9}^{0} y + 9 d y

x \to 3 lim (x^{2} + 8 x - 2)

n = 5 \sum \infty \frac{2}{n ^{2} - 1}

What is the integral of 8/(x+1)+x ?
The integral of 8/(x+1)+x is 8ln|x+1|+(x^2)/2+C