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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

Apply Integration By Parts

= 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

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