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

The integral of 8(x+1)ln(x+1) is 8(1/2 x^2ln(x+1)+xln(x+1)+1/2 ln(x+1)-1/4 (x+1)^2)+C

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

\int 8 (x + 1) ln (x + 1) d x

8 (\frac{1}{2} x^{2} ln (x + 1) + x ln (x + 1) + \frac{1}{2} ln (x + 1) - \frac{1}{4} (x + 1)^{2}) + C

Solution steps

\int 8 (x + 1) ln (x + 1) d x

Take the constant out:

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

= 8 \cdot \int (x + 1) ln (x + 1) d x

Apply u-substitution

= 8 \cdot \int u ln (u) d u

Apply Integration By Parts

= 8 (\frac{1}{2} u^{2} ln (u) - \int \frac{u}{2} d u)

\int \frac{u}{2} d u = \frac{u ^{2}}{4}

= 8 (\frac{1}{2} u^{2} ln (u) - \frac{u ^{2}}{4})

Substitute back

u = x + 1

= 8 (\frac{1}{2} (x + 1)^{2} ln (x + 1) - \frac{( x + 1 ) ^{2}}{4})

Simplify

8 (\frac{1}{2} (x + 1)^{2} ln (x + 1) - \frac{( x + 1 ) ^{2}}{4}) : 8 (\frac{1}{2} x^{2} ln (x + 1) + x ln (x + 1) + \frac{1}{2} ln (x + 1) - \frac{1}{4} (x + 1)^{2})

= 8 (\frac{1}{2} x^{2} ln (x + 1) + x ln (x + 1) + \frac{1}{2} ln (x + 1) - \frac{1}{4} (x + 1)^{2})

Add a constant to the solution

= 8 (\frac{1}{2} x^{2} ln (x + 1) + x ln (x + 1) + \frac{1}{2} ln (x + 1) - \frac{1}{4} (x + 1)^{2}) + C

What is the integral of 8(x+1)ln(x+1) ?
The integral of 8(x+1)ln(x+1) is 8(1/2 x^2ln(x+1)+xln(x+1)+1/2 ln(x+1)-1/4 (x+1)^2)+C