What is the integral of tln(t+4) ?

The integral of tln(t+4) is ln(t+4)(1/2 (t+4)^2-4(t+4))-1/4 (t^2-8t-48)+C

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integral of tln(t+4)

\int t ln (t + 4) d t

ln (t + 4) (\frac{1}{2} (t + 4)^{2} - 4 (t + 4)) - \frac{1}{4} (t^{2} - 8 t - 48) + C

Solution steps

\int t ln (t + 4) d t

Apply u-substitution

: \int (u - 4) ln (u) d u

= \int (u - 4) ln (u) d u

Apply Integration By Parts

: ln (u) (\frac{u ^{2}}{2} - 4 u) - \int \frac{1}{2} (u - 8) d u

= ln (u) (\frac{u ^{2}}{2} - 4 u) - \int \frac{1}{2} (u - 8) d u

\int \frac{1}{2} (u - 8) d u = \frac{1}{2} (\frac{u ^{2}}{2} - 8 u)

= ln (u) (\frac{u ^{2}}{2} - 4 u) - \frac{1}{2} (\frac{u ^{2}}{2} - 8 u)

Substitute back

u = t + 4

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

Simplify

ln (t + 4) (\frac{( t + 4 ) ^{2}}{2} - 4 (t + 4)) - \frac{1}{2} (\frac{( t + 4 ) ^{2}}{2} - 8 (t + 4)) : ln (t + 4) (\frac{1}{2} (t + 4)^{2} - 4 (t + 4)) - \frac{1}{4} (t^{2} - 8 t - 48)

= ln (t + 4) (\frac{1}{2} (t + 4)^{2} - 4 (t + 4)) - \frac{1}{4} (t^{2} - 8 t - 48)

Add a constant to the solution

= ln (t + 4) (\frac{1}{2} (t + 4)^{2} - 4 (t + 4)) - \frac{1}{4} (t^{2} - 8 t - 48) + C

\frac{d}{d x} (\frac{36 x}{x ^{2} + 36})

f (x) = (4 + 5 x)^{2}, (- 3, 121)

f (x) = 2 x, (18, 6)

\frac{d y}{d x} = \frac{e ^{y} + 1}{e ^{y}}

x = - 2, x = 1, y = 2 x^{2} + 7, y = 0

What is the integral of tln(t+4) ?
The integral of tln(t+4) is ln(t+4)(1/2 (t+4)^2-4(t+4))-1/4 (t^2-8t-48)+C