What is the integral of tln(t+3) ?

The integral of tln(t+3) is ln(t+3)(1/2 (t+3)^2-3(t+3))-1/4 (t^2-6t-27)+C

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integral of tln(t+3)

\int t ln (t + 3) d t

ln (t + 3) (\frac{1}{2} (t + 3)^{2} - 3 (t + 3)) - \frac{1}{4} (t^{2} - 6 t - 27) + C

Solution steps

\int t ln (t + 3) d t

Apply u-substitution

= \int (u - 3) ln (u) d u

Apply Integration By Parts

= ln (u) (\frac{u ^{2}}{2} - 3 u) - \int \frac{1}{2} (u - 6) d u

\int \frac{1}{2} (u - 6) d u = \frac{1}{2} (\frac{u ^{2}}{2} - 6 u)

= ln (u) (\frac{u ^{2}}{2} - 3 u) - \frac{1}{2} (\frac{u ^{2}}{2} - 6 u)

Substitute back

u = t + 3

= ln (t + 3) (\frac{( t + 3 ) ^{2}}{2} - 3 (t + 3)) - \frac{1}{2} (\frac{( t + 3 ) ^{2}}{2} - 6 (t + 3))

Simplify

ln (t + 3) (\frac{( t + 3 ) ^{2}}{2} - 3 (t + 3)) - \frac{1}{2} (\frac{( t + 3 ) ^{2}}{2} - 6 (t + 3)) : ln (t + 3) (\frac{1}{2} (t + 3)^{2} - 3 (t + 3)) - \frac{1}{4} (t^{2} - 6 t - 27)

= ln (t + 3) (\frac{1}{2} (t + 3)^{2} - 3 (t + 3)) - \frac{1}{4} (t^{2} - 6 t - 27)

Add a constant to the solution

= ln (t + 3) (\frac{1}{2} (t + 3)^{2} - 3 (t + 3)) - \frac{1}{4} (t^{2} - 6 t - 27) + C

What is the integral of tln(t+3) ?
The integral of tln(t+3) is ln(t+3)(1/2 (t+3)^2-3(t+3))-1/4 (t^2-6t-27)+C