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

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

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integral of ln(1+t)

\int ln (1 + t) d t

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

Solution steps

\int ln (1 + t) 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 = 1 + t

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

Simplify

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

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

Add a constant to the solution

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

\int e^{- \frac{x}{100}} d x

\int 15 x^{2} e^{5 x^{3} - 6} d x

s \to 0 lim (\frac{- \frac{1}{s}}{1 + ( 3 + \frac{13}{s} ) ( \frac{7}{x s + 1} )})

x \to 1 + lim (\frac{3}{∣ x - 1 ∣})

x^{3}

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