What is the integral of (2x-3)/(x^2+4) ?

The integral of (2x-3)/(x^2+4) is ln|(x^2)/4+1|-3/2 arctan(x/2)+C

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integral of (2x-3)/(x^2+4)

\int \frac{2 x - 3}{x ^{2} + 4} d x

ln \frac{x ^{2}}{4} + 1 - \frac{3}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{2 x - 3}{x ^{2} + 4} d x

Apply Integral Substitution

= \int \frac{4 u - 3}{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{4 u - 3}{u ^{2} + 1} d u

Expand

\frac{4 u - 3}{u ^{2} + 1} : \frac{4 u}{u ^{2} + 1} - \frac{3}{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{4 u}{u ^{2} + 1} d u - \int \frac{3}{u ^{2} + 1} d u)

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

\int \frac{3}{u ^{2} + 1} d u = 3 arctan (u)

= \frac{1}{2} (2 ln u^{2} + 1 - 3 arctan (u))

Substitute back

u = \frac{x}{2}

= \frac{1}{2} (2 ln (\frac{x}{2})^{2} + 1 - 3 arctan (\frac{x}{2}))

Simplify

\frac{1}{2} (2 ln (\frac{x}{2})^{2} + 1 - 3 arctan (\frac{x}{2})) : ln \frac{x ^{2}}{4} + 1 - \frac{3}{2} arctan (\frac{x}{2})

= ln \frac{x ^{2}}{4} + 1 - \frac{3}{2} arctan (\frac{x}{2})

Add a constant to the solution

= ln \frac{x ^{2}}{4} + 1 - \frac{3}{2} arctan (\frac{x}{2}) + C

n = 1 \sum \infty \frac{2 n}{n ^{2}}

f (x) = 7 arctan (7 x)

y = - 8 x ln (6 x + 5)

\int_{0}^{1} (x + 3) (x - 4) d x

y = (x^{3} - 16 x)^{12}, (4, 0)

What is the integral of (2x-3)/(x^2+4) ?
The integral of (2x-3)/(x^2+4) is ln|(x^2)/4+1|-3/2 arctan(x/2)+C