What is the integral of 15xln(1+x) ?

The integral of 15xln(1+x) is 15(1/2 ln(1+x)(1+x)^2-ln(1+x)-xln(1+x)-1/4 (1+x)^2+1+x)+C

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integral of 15xln(1+x)

\int 15 x ln (1 + x) d x

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

Solution steps

\int 15 x ln (1 + x) d x

Take the constant out:

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

= 15 \cdot \int x ln (1 + x) d x

Apply u-substitution

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

= 15 \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

= 15 (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

= 15 (ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u))

Substitute back

u = 1 + x

= 15 (ln (1 + x) (\frac{( 1 + x ) ^{2}}{2} - (1 + x)) - (\frac{( 1 + x ) ^{2}}{4} - (1 + x)))

Simplify

15 (ln (1 + x) (\frac{( 1 + x ) ^{2}}{2} - (1 + x)) - (\frac{( 1 + x ) ^{2}}{4} - (1 + x))) : 15 (\frac{1}{2} ln (1 + x) (1 + x)^{2} - ln (1 + x) - x ln (1 + x) - \frac{1}{4} (1 + x)^{2} + 1 + x)

= 15 (\frac{1}{2} ln (1 + x) (1 + x)^{2} - ln (1 + x) - x ln (1 + x) - \frac{1}{4} (1 + x)^{2} + 1 + x)

Add a constant to the solution

= 15 (\frac{1}{2} ln (1 + x) (1 + x)^{2} - ln (1 + x) - x ln (1 + x) - \frac{1}{4} (1 + x)^{2} + 1 + x) + C

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What is the integral of 15xln(1+x) ?
The integral of 15xln(1+x) is 15(1/2 ln(1+x)(1+x)^2-ln(1+x)-xln(1+x)-1/4 (1+x)^2+1+x)+C