What is the integral of (6x-5)/(x^2+100) ?

The integral of (6x-5)/(x^2+100) is 1/2 (6ln|(x^2)/(100)+1|-arctan(x/(10)))+C

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integral of (6x-5)/(x^2+100)

\int \frac{6 x - 5}{x ^{2} + 100} d x

\frac{1}{2} (6 ln \frac{x ^{2}}{100} + 1 - arctan (\frac{x}{10})) + C

Solution steps

\int \frac{6 x - 5}{x ^{2} + 100} d x

Apply Integral Substitution

= \int \frac{12 u - 1}{2 ( 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}{2} \cdot \int \frac{12 u - 1}{u ^{2} + 1} d u

Expand

\frac{12 u - 1}{u ^{2} + 1} : \frac{12 u}{u ^{2} + 1} - \frac{1}{u ^{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{12 u}{u ^{2} + 1} d u - \int \frac{1}{u ^{2} + 1} d u)

\int \frac{12 u}{u ^{2} + 1} d u = 6 ln u^{2} + 1

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

= \frac{1}{2} (6 ln u^{2} + 1 - arctan (u))

Substitute back

u = \frac{x}{10}

= \frac{1}{2} (6 ln (\frac{x}{10})^{2} + 1 - arctan (\frac{x}{10}))

(\frac{x}{10})^{2} = \frac{x ^{2}}{100}

= \frac{1}{2} (6 ln \frac{x ^{2}}{100} + 1 - arctan (\frac{x}{10}))

Add a constant to the solution

= \frac{1}{2} (6 ln \frac{x ^{2}}{100} + 1 - arctan (\frac{x}{10})) + C

What is the integral of (6x-5)/(x^2+100) ?
The integral of (6x-5)/(x^2+100) is 1/2 (6ln|(x^2)/(100)+1|-arctan(x/(10)))+C