What is the integral of tln(1+t) ?

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

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integral of tln(1+t)

\int t ln (1 + t) d t

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

Solution steps

\int t ln (1 + t) 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 = 1 + t

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

Simplify

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

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

Add a constant to the solution

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

\int_{- 8}^{0} (\frac{y}{8} + y + 9) d y

x \to \infty lim (x + 9)

\int \frac{1}{x ^{2} + 3 x + 3} d x

y^{''} (x) - 3 y^{'} (x) + 2 y = \frac{1}{1 + e ^{- x}}

x \to 0 lim (- arcsin (x))

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