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

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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Expand

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

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

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

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

Substitute back

u = \frac{x}{2}

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

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

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

Add a constant to the solution

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

(x + 2)^{3}, 2 x^{2} - 8 x + 8, 0

x \to 1 lim (x + 4 - 12)

f (x) = x + 2, at x = 1

f (x) = lo g_{4} (3 x^{2} + 1)

x \to 14 - lim (15 x + ln (14 - x))

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