What is the integral from 0 to x^2 of ln(1+t) ?

The integral from 0 to x^2 of ln(1+t) is ln(x^2+1)+x^2ln(x^2+1)-x^2

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integral from 0 to x^2 of ln(1+t)

\int_{0}^{x^{2}} ln (1 + t) d t

ln (x^{2} + 1) + x^{2} ln (x^{2} + 1) - x^{2}

Solution steps

\int_{0}^{x^{2}} ln (1 + t) d t

Apply u-substitution

: \int_{1}^{1 + x^{2}} ln (u) d u

= \int_{1}^{1 + x^{2}} ln (u) d u

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{1}^{1 + x^{2}}

= [u ln (u) - \int 1 d u]_{1}^{1 + x^{2}}

\int 1 d u = u

= [u ln (u) - u]_{1}^{1 + x^{2}}

Compute the boundaries:

ln (x^{2} + 1) + x^{2} ln (x^{2} + 1) - x^{2}

= ln (x^{2} + 1) + x^{2} ln (x^{2} + 1) - x^{2}

\int_{- 2}^{0} \frac{1}{e ^{- 2 x}} d x

\int_{0}^{1} e^{x} + 1 d x

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

\int_{0}^{2 π} 2 d x

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

What is the integral from 0 to x^2 of ln(1+t) ?
The integral from 0 to x^2 of ln(1+t) is ln(x^2+1)+x^2ln(x^2+1)-x^2