What is the integral of (2x-5)/(x^2+8) ?

The integral of (2x-5)/(x^2+8) is ln|(x^2)/8+1|-5/(2sqrt(2))arctan(x/(2sqrt(2)))+C

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integral of (2x-5)/(x^2+8)

\int \frac{2 x - 5}{x ^{2} + 8} d x

ln \frac{x ^{2}}{8} + 1 - \frac{5}{2 2} arctan (\frac{x}{2 2}) + C

Solution steps

\int \frac{2 x - 5}{x ^{2} + 8} d x

Apply Integral Substitution

= \int \frac{4 2 u - 5}{2 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 2} \cdot \int \frac{4 2 u - 5}{u ^{2} + 1} d u

Expand

\frac{4 2 u - 5}{u ^{2} + 1} : \frac{4 2 u}{u ^{2} + 1} - \frac{5}{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 2} (\int \frac{4 2 u}{u ^{2} + 1} d u - \int \frac{5}{u ^{2} + 1} d u)

\int \frac{4 2 u}{u ^{2} + 1} d u = 22 ln u^{2} + 1

\int \frac{5}{u ^{2} + 1} d u = 5 arctan (u)

= \frac{1}{2 2} (22 ln u^{2} + 1 - 5 arctan (u))

Substitute back

u = \frac{x}{2 2}

= \frac{1}{2 2} (22 ln (\frac{x}{2 2})^{2} + 1 - 5 arctan (\frac{x}{2 2}))

Simplify

\frac{1}{2 2} (22 ln (\frac{x}{2 2})^{2} + 1 - 5 arctan (\frac{x}{2 2})) : ln \frac{x ^{2}}{8} + 1 - \frac{5}{2 2} arctan (\frac{x}{2 2})

= ln \frac{x ^{2}}{8} + 1 - \frac{5}{2 2} arctan (\frac{x}{2 2})

Add a constant to the solution

= ln \frac{x ^{2}}{8} + 1 - \frac{5}{2 2} arctan (\frac{x}{2 2}) + C

\frac{d}{d x} (sinh (x^{2}))

f (x) = \frac{x ^{2} + x + 1}{x ^{2} + 1}

\frac{1}{2 - x}

g (x) = \frac{- 7}{x}

x \to - 2 lim (3 x^{3} - 2 x + 7)

What is the integral of (2x-5)/(x^2+8) ?
The integral of (2x-5)/(x^2+8) is ln|(x^2)/8+1|-5/(2sqrt(2))arctan(x/(2sqrt(2)))+C