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

The integral of tln(sqrt(t+1)) is 1/8 (2t^2ln(t+1)-t^2+2t-2ln(t+1)+3)+C

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integral of tln(sqrt(t+1))

\int t ln (t + 1) d t

\frac{1}{8} (2 t^{2} ln (t + 1) - t^{2} + 2 t - 2 ln (t + 1) + 3) + C

Solution steps

\int t ln (t + 1) d t

Simplify

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

= \int \frac{1}{2} t ln (t + 1) d t

Take the constant out:

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

= \frac{1}{2} \cdot \int t ln (t + 1) d t

Apply u-substitution

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

= \frac{1}{2} \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

= \frac{1}{2} (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

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

Substitute back

u = t + 1

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

Simplify

\frac{1}{2} (ln (t + 1) (\frac{( t + 1 ) ^{2}}{2} - (t + 1)) - (\frac{( t + 1 ) ^{2}}{4} - (t + 1))) : \frac{1}{8} (2 t^{2} ln (t + 1) - t^{2} + 2 t - 2 ln (t + 1) + 3)

= \frac{1}{8} (2 t^{2} ln (t + 1) - t^{2} + 2 t - 2 ln (t + 1) + 3)

Add a constant to the solution

= \frac{1}{8} (2 t^{2} ln (t + 1) - t^{2} + 2 t - 2 ln (t + 1) + 3) + C

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What is the integral of tln(sqrt(t+1)) ?
The integral of tln(sqrt(t+1)) is 1/8 (2t^2ln(t+1)-t^2+2t-2ln(t+1)+3)+C