What is the integral from 0 to x of ln(1+x) ?

The integral from 0 to x of ln(1+x) is xln(x+1)-x+ln(x+1)

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integral from 0 to x of ln(1+x)

\int_{0}^{x} ln (1 + x) d x

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

Solution steps

\int_{0}^{x} ln (1 + x) d x

Apply u-substitution

: \int_{1}^{1 + x} ln (u) d u

= \int_{1}^{1 + x} ln (u) d u

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{1}^{1 + x}

= [u ln (u) - \int 1 d u]_{1}^{1 + x}

\int 1 d u = u

= [u ln (u) - u]_{1}^{1 + x}

Compute the boundaries:

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

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

\int_{0}^{0.3} x ln (1 + x^{3}) d x

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

\int_{- 2}^{2} x^{5} d x

\int_{2}^{x^{3}} \frac{1}{t ^{3}} d t

\int_{0}^{5} 6

What is the integral from 0 to x of ln(1+x) ?
The integral from 0 to x of ln(1+x) is xln(x+1)-x+ln(x+1)