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

The integral of 8(x+1)In(x+1) is 8In(x+x^2+(x^3)/3)+C

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integral of 8(x+1)In(x+1)

\int 8 (x + 1) I n (x + 1) d x

8 I n (x + x^{2} + \frac{x ^{3}}{3}) + C

Solution steps

\int 8 (x + 1) I n (x + 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 I n \cdot \int (1 + x) (1 + x) d x

Simplify

(1 + x) (1 + x) : (1 + x)^{2}

= 8 I n \cdot \int (1 + x)^{2} d x

Expand

(1 + x)^{2} : 1 + 2 x + x^{2}

= 8 I n \cdot \int 1 + 2 x + x^{2} 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

= 8 I n (\int 1 d x + \int 2 x d x + \int x^{2} d x)

\int 1 d x = x

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

\int x^{2} d x = \frac{x ^{3}}{3}

= 8 I n (x + x^{2} + \frac{x ^{3}}{3})

Add a constant to the solution

= 8 I n (x + x^{2} + \frac{x ^{3}}{3}) + C

n = 0 \sum \infty z^{3 n}

x \to \infty lim (\frac{a _{n + 1}}{a _{n}})

\frac{d r}{d t} = \frac{1}{t + 1}, r (0) = 1

(1 + x^{2}) y^{^{'}} + 1 + 2 x y = 0

\int x^{2} (3 - e^{x^{3}}) d x

What is the integral of 8(x+1)In(x+1) ?
The integral of 8(x+1)In(x+1) is 8In(x+x^2+(x^3)/3)+C