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

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

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

\int \frac{( x + 2 )}{x ^{2} + 4} d x

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

Solution steps

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

Apply Integral Substitution

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

Expand

\frac{u + 1}{u ^{2} + 1} : \frac{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

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

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

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

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

Substitute back

u = \frac{x}{2}

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

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

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

Add a constant to the solution

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

- 4 x^{2} y + y^{^{'}} = - 10 x^{2}

\int \frac{10 x}{x ^{2} - 2 x - 24} d x

f (x) = \frac{e ^{x}}{x}

8 t

\frac{d ^{2}}{d x ^{2}} (x + 7)

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