What is the integral of (5x-6)/(x^2+4) ?

The integral of (5x-6)/(x^2+4) is 5/2 ln|(x^2)/4+1|-3arctan(x/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5x-6)/(x^2+4)

\int \frac{5 x - 6}{x ^{2} + 4} d x

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

Solution steps

\int \frac{5 x - 6}{x ^{2} + 4} d x

Apply Integral Substitution

= \int \frac{5 u - 3}{u ^{2} + 1} d u

Expand

\frac{5 u - 3}{u ^{2} + 1} : \frac{5 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

= \int \frac{5 u}{u ^{2} + 1} d u - \int \frac{3}{u ^{2} + 1} d u

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

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

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

Substitute back

u = \frac{x}{2}

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (sin (7 x^{4} y - 7 x y^{4}))

y^{^{''}} - 10 y^{^{'}} + 34 y = 0

\frac{\partial}{\partial x} (x^{- \frac{1}{2}} cos (x))

\frac{s - 1}{s ( s ^{2} + s + 1 )}

\frac{\partial}{\partial x} (e^{e^{x y}})

What is the integral of (5x-6)/(x^2+4) ?
The integral of (5x-6)/(x^2+4) is 5/2 ln|(x^2)/4+1|-3arctan(x/2)+C