What is the integral of tln(t+2) ?

The integral of tln(t+2) is ln(t+2)(1/2 (t+2)^2-2(t+2))-1/4 (t^2-4t-12)+C

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integral of tln(t+2)

\int t ln (t + 2) d t

ln (t + 2) (\frac{1}{2} (t + 2)^{2} - 2 (t + 2)) - \frac{1}{4} (t^{2} - 4 t - 12) + C

Solution steps

\int t ln (t + 2) d t

Apply u-substitution

: \int (u - 2) ln (u) d u

= \int (u - 2) ln (u) d u

Apply Integration By Parts

: ln (u) (\frac{u ^{2}}{2} - 2 u) - \int \frac{1}{2} (u - 4) d u

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

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

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

Substitute back

u = t + 2

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

Simplify

ln (t + 2) (\frac{( t + 2 ) ^{2}}{2} - 2 (t + 2)) - \frac{1}{2} (\frac{( t + 2 ) ^{2}}{2} - 4 (t + 2)) : ln (t + 2) (\frac{1}{2} (t + 2)^{2} - 2 (t + 2)) - \frac{1}{4} (t^{2} - 4 t - 12)

= ln (t + 2) (\frac{1}{2} (t + 2)^{2} - 2 (t + 2)) - \frac{1}{4} (t^{2} - 4 t - 12)

Add a constant to the solution

= ln (t + 2) (\frac{1}{2} (t + 2)^{2} - 2 (t + 2)) - \frac{1}{4} (t^{2} - 4 t - 12) + C

\frac{d}{d t} (sin (5 t))

y^{^{'}} = 2 x - 3 y

\int 2 (tan^{2} (x) + tan^{4} (x)) d x

\int x arccot (x) d x

f^{^{''}} (x) = x^{- \frac{3}{2}}, f^{^{'}} (4) = 7, f (0) = 0

What is the integral of tln(t+2) ?
The integral of tln(t+2) is ln(t+2)(1/2 (t+2)^2-2(t+2))-1/4 (t^2-4t-12)+C