What is the integral of (u-2)/(u^2+16) ?

The integral of (u-2)/(u^2+16) is 1/2 (ln|(u^2)/(16)+1|-arctan(u/4))+C

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integral of (u-2)/(u^2+16)

\int \frac{u - 2}{u ^{2} + 16} d u

\frac{1}{2} (ln \frac{u ^{2}}{16} + 1 - arctan (\frac{u}{4})) + C

Solution steps

\int \frac{u - 2}{u ^{2} + 16} d u

Apply Integral Substitution

= \int \frac{2 v - 1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{2 v - 1}{v ^{2} + 1} d v

Expand

\frac{2 v - 1}{v ^{2} + 1} : \frac{2 v}{v ^{2} + 1} - \frac{1}{v ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{2 v}{v ^{2} + 1} d v - \int \frac{1}{v ^{2} + 1} d v)

\int \frac{2 v}{v ^{2} + 1} d v = ln v^{2} + 1

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

= \frac{1}{2} (ln v^{2} + 1 - arctan (v))

Substitute back

v = \frac{u}{4}

= \frac{1}{2} (ln (\frac{u}{4})^{2} + 1 - arctan (\frac{u}{4}))

(\frac{u}{4})^{2} = \frac{u ^{2}}{16}

= \frac{1}{2} (ln \frac{u ^{2}}{16} + 1 - arctan (\frac{u}{4}))

Add a constant to the solution

= \frac{1}{2} (ln \frac{u ^{2}}{16} + 1 - arctan (\frac{u}{4})) + C

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n = 0 \sum \infty \frac{1}{4 n}

What is the integral of (u-2)/(u^2+16) ?
The integral of (u-2)/(u^2+16) is 1/2 (ln|(u^2)/(16)+1|-arctan(u/4))+C