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

The integral from 0 to x of 1/(x^2+y^2) is pi/(4x)

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

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

\frac{π}{4 x}

Solution steps

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

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{x ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{x} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{x} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{x} \cdot \frac{π}{4}

Simplify

= \frac{π}{4 x}

\int_{- 2}^{2} 4 - y^{2} d y

\int_{0}^{1} - x^{2} + 1 d x

\int_{- 1}^{1} \frac{1}{5 - 4 x} d x

\int_{2}^{4} 2 d x

\int_{1}^{2} (1 + \frac{1}{u} + \frac{1}{u ^{2}}) d u

What is the integral from 0 to x of 1/(x^2+y^2) ?
The integral from 0 to x of 1/(x^2+y^2) is pi/(4x)