What is the integral of 10xln(x+1) ?

The integral of 10xln(x+1) is 10(1/2 ln(x+1)(x+1)^2-xln(x+1)-ln(x+1)-1/4 (x+1)^2+x+1)+C

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integral of 10xln(x+1)

\int 10 x ln (x + 1) d x

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

Solution steps

\int 10 x ln (x + 1) d x

Take the constant out:

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

= 10 \cdot \int x ln (x + 1) d x

Apply u-substitution

: \int (u - 1) ln (u) d u

= 10 \cdot \int (u - 1) ln (u) d u

Apply Integration By Parts

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

= 10 (ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u)

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= 10 (ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u))

Substitute back

u = x + 1

= 10 (ln (x + 1) (\frac{( x + 1 ) ^{2}}{2} - (x + 1)) - (\frac{( x + 1 ) ^{2}}{4} - (x + 1)))

Simplify

10 (ln (x + 1) (\frac{( x + 1 ) ^{2}}{2} - (x + 1)) - (\frac{( x + 1 ) ^{2}}{4} - (x + 1))) : 10 (\frac{1}{2} ln (x + 1) (x + 1)^{2} - x ln (x + 1) - ln (x + 1) - \frac{1}{4} (x + 1)^{2} + x + 1)

= 10 (\frac{1}{2} ln (x + 1) (x + 1)^{2} - x ln (x + 1) - ln (x + 1) - \frac{1}{4} (x + 1)^{2} + x + 1)

Add a constant to the solution

= 10 (\frac{1}{2} ln (x + 1) (x + 1)^{2} - x ln (x + 1) - ln (x + 1) - \frac{1}{4} (x + 1)^{2} + x + 1) + C

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\frac{\partial}{\partial x} (4 (x y)^{2})

What is the integral of 10xln(x+1) ?
The integral of 10xln(x+1) is 10(1/2 ln(x+1)(x+1)^2-xln(x+1)-ln(x+1)-1/4 (x+1)^2+x+1)+C