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

The integral of ln(1+x) is ln(1+x)+xln(1+x)-1-x+C

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integral of ln(1+x)

\int ln (1 + x) d x

ln (1 + x) + x ln (1 + x) - 1 - x + C

Solution steps

\int ln (1 + x) d x

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 + x

= (1 + x) ln (1 + x) - (1 + x)

Simplify

(1 + x) ln (1 + x) - (1 + x) : ln (1 + x) + x ln (1 + x) - 1 - x

= ln (1 + x) + x ln (1 + x) - 1 - x

Add a constant to the solution

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

\frac{d}{d y} (\frac{y - 1}{y ^{2} - y + 1})

\int_{1}^{2} 2 (x^{2} - 4) d x

\int \frac{y + 13}{( y + 1 ) ( y + 3 ) ( y - 2 )} d y

\int \frac{1}{- t ^{2} + 8 t - 12} d t

(e^{x} + e^{- x}) (\frac{d y}{d x}) = y^{2}

What is the integral of ln(1+x) ?
The integral of ln(1+x) is ln(1+x)+xln(1+x)-1-x+C