What is the integral from y to 8-y of 3/(x^2+y^2) ?

The integral from y to 8-y of 3/(x^2+y^2) is (3(arctan((8-y)/x)-arctan(y/x)))/(x)

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integral from y to 8-y of 3/(x^2+y^2)

\int_{y}^{8 - y} \frac{3}{x ^{2} + y ^{2}} d y

\frac{3 ( arctan ( \frac{8 - y}{x} ) - arctan ( \frac{y}{x} ) )}{x}

Solution steps

\int_{y}^{8 - y} \frac{3}{x ^{2} + y ^{2}} d y

Take the constant out:

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

= 3 \cdot \int_{y}^{8 - y} \frac{1}{x ^{2} + y ^{2}} d y

Apply Integral Substitution

= 3 \cdot \int_{\frac{y}{x}}^{\frac{8 - y}{x}} \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

= 3 \cdot \frac{1}{x} \cdot \int_{\frac{y}{x}}^{\frac{8 - y}{x}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 3 \cdot \frac{1}{x} [arctan (u)]_{\frac{y}{x}}^{\frac{8 - y}{x}}

Simplify

3 \cdot \frac{1}{x} [arctan (u)]_{\frac{y}{x}}^{\frac{8 - y}{x}} : \frac{3}{x} [arctan (u)]_{\frac{y}{x}}^{\frac{8 - y}{x}}

= \frac{3}{x} [arctan (u)]_{\frac{y}{x}}^{\frac{8 - y}{x}}

Compute the boundaries:

arctan (\frac{8 - y}{x}) - arctan (\frac{y}{x})

= \frac{3}{x} (arctan (\frac{8 - y}{x}) - arctan (\frac{y}{x}))

Simplify

= \frac{3 ( arctan ( \frac{8 - y}{x} ) - arctan ( \frac{y}{x} ) )}{x}

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What is the integral from y to 8-y of 3/(x^2+y^2) ?
The integral from y to 8-y of 3/(x^2+y^2) is (3(arctan((8-y)/x)-arctan(y/x)))/(x)