What is the integral of (5x-3)/(x^2+4) ?

The integral of (5x-3)/(x^2+4) is 1/2 (5ln|(x^2)/4+1|-3arctan(x/2))+C

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integral of (5x-3)/(x^2+4)

\int \frac{5 x - 3}{x ^{2} + 4} d x

\frac{1}{2} (5 ln \frac{x ^{2}}{4} + 1 - 3 arctan (\frac{x}{2})) + C

Solution steps

\int \frac{5 x - 3}{x ^{2} + 4} d x

Apply Integral Substitution

= \int \frac{10 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{10 u - 3}{u ^{2} + 1} d u

Expand

\frac{10 u - 3}{u ^{2} + 1} : \frac{10 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{10 u}{u ^{2} + 1} d u - \int \frac{3}{u ^{2} + 1} d u)

\int \frac{10 u}{u ^{2} + 1} d u = 5 ln u^{2} + 1

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

= \frac{1}{2} (5 ln u^{2} + 1 - 3 arctan (u))

Substitute back

u = \frac{x}{2}

= \frac{1}{2} (5 ln (\frac{x}{2})^{2} + 1 - 3 arctan (\frac{x}{2}))

(\frac{x}{2})^{2} = \frac{x ^{2}}{4}

= \frac{1}{2} (5 ln \frac{x ^{2}}{4} + 1 - 3 arctan (\frac{x}{2}))

Add a constant to the solution

= \frac{1}{2} (5 ln \frac{x ^{2}}{4} + 1 - 3 arctan (\frac{x}{2})) + C

\int \frac{1}{x ^{2} + 4 x + 13} d x

x \to π lim (ln ∣ cos (x) - 7 ∣)

\int_{0}^{\frac{π}{2}} cos^{7} (x) d x

6 y^{^{''}} + 10 y^{^{'}} = 0

\frac{d y}{d t}

What is the integral of (5x-3)/(x^2+4) ?
The integral of (5x-3)/(x^2+4) is 1/2 (5ln|(x^2)/4+1|-3arctan(x/2))+C