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

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

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integral of ln(x+1)

\int ln (x + 1) d x

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{( 12 + 4 x )}{x - 1}

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\frac{\partial}{\partial x} ((x^{3} + y^{2})^{5})

\int_{0}^{\frac{π}{2}} 20 cos^{5} (x) d x

\int_{1}^{2} (\frac{6}{x ^{3}} + 2 x) d x

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