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

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

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integral from 0 to x^2 of 1/(x+y)

\int_{0}^{x^{2}} \frac{1}{x + y} d y

ln x + x^{2} - ln ∣ x ∣

Solution steps

\int_{0}^{x^{2}} \frac{1}{x + y} d y

Apply u-substitution

: \int_{x}^{x + x^{2}} \frac{1}{u} d u

= \int_{x}^{x + x^{2}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{x}^{x + x^{2}}

Compute the boundaries:

ln x + x^{2} - ln ∣ x ∣

= ln x + x^{2} - ln ∣ x ∣

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