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

The integral of ln(t+1) is tln(t+1)+ln(t+1)-t-1+C

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integral of ln(t+1)

\int ln (t + 1) d t

t ln (t + 1) + ln (t + 1) - t - 1 + C

Solution steps

\int ln (t + 1) d t

Apply u-substitution

: \int ln (u) d u

= \int ln (u) d u

Apply Integration By Parts

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

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

\int 1 d u = u

= u ln (u) - u

Substitute back

u = t + 1

= (t + 1) ln (t + 1) - (t + 1)

Simplify

(t + 1) ln (t + 1) - (t + 1) : t ln (t + 1) + ln (t + 1) - t - 1

= t ln (t + 1) + ln (t + 1) - t - 1

Add a constant to the solution

= t ln (t + 1) + ln (t + 1) - t - 1 + C

\int \frac{- 5}{4 - 2 x - x ^{2}} d x

\frac{d}{d x} (\frac{sin ( x )}{x ^{5}})

\frac{d}{d x} (x^{4} - 2 x^{2} + 3)

x^{3} - 2 x + 1

y = (x^{3} - 9 x)^{10}, (3, 0)

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