What is the integral of t*ln(t+1) ?

The integral of t*ln(t+1) is 1/2 ln(t+1)(t+1)^2-tln(t+1)-ln(t+1)-1/4 (t+1)^2+t+1+C

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integral of t*ln(t+1)

\int t \cdot ln (t + 1) d t

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

Solution steps

\int t ln (t + 1) d t

Apply u-substitution

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

= \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

= 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

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

Substitute back

u = t + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{3}{x + 9})

- 7 z^{4} + z^{2} - 25

\frac{\partial}{\partial x} (\frac{( x ^{2} )}{y + 3})

x \to 4 lim (\frac{∣ x ∣ - 4}{x - 4})

\frac{\partial}{\partial z} (x e^{- y} - z)

What is the integral of t*ln(t+1) ?
The integral of t*ln(t+1) is 1/2 ln(t+1)(t+1)^2-tln(t+1)-ln(t+1)-1/4 (t+1)^2+t+1+C