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

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

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

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

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

Solution steps

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

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:

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

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

\int_{x}^{x^{2}} (\frac{y}{x}) d y

\int_{0}^{6} \frac{∣ 2 x - 5 ∣}{x + 1} d x

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

\int_{0}^{6} x d x

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

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