What is the integral of 6xln(x+1) ?

The integral of 6xln(x+1) is 6(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 6xln(x+1)

\int 6 x ln (x + 1) d x

6 (\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 6 x ln (x + 1) d x

Take the constant out:

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

= 6 \cdot \int x ln (x + 1) d x

Apply u-substitution

= 6 \cdot \int (u - 1) ln (u) d u

Apply Integration By Parts

= 6 (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

= 6 (ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u))

Substitute back

u = x + 1

= 6 (ln (x + 1) (\frac{( x + 1 ) ^{2}}{2} - (x + 1)) - (\frac{( x + 1 ) ^{2}}{4} - (x + 1)))

Simplify

6 (ln (x + 1) (\frac{( x + 1 ) ^{2}}{2} - (x + 1)) - (\frac{( x + 1 ) ^{2}}{4} - (x + 1))) : 6 (\frac{1}{2} ln (x + 1) (x + 1)^{2} - x ln (x + 1) - ln (x + 1) - \frac{1}{4} (x + 1)^{2} + x + 1)

= 6 (\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

= 6 (\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

\frac{d}{d x} (24 x)

\int \frac{1 + 2 x ^{2}}{x} d x

\int \frac{144}{x ^{3} - 12 x ^{2}} d x

f (x) = 5 cos (x), at x = \frac{π}{3}

cos (x), - cos (x) + 2

What is the integral of 6xln(x+1) ?
The integral of 6xln(x+1) is 6(1/2 ln(x+1)(x+1)^2-xln(x+1)-ln(x+1)-1/4 (x+1)^2+x+1)+C