What is the integral of (x^2+9)/(x^2+4) ?

The integral of (x^2+9)/(x^2+4) is 5/2 arctan(x/2)+x+C

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integral of (x^2+9)/(x^2+4)

\int \frac{x ^{2} + 9}{x ^{2} + 4} d x

\frac{5}{2} arctan (\frac{x}{2}) + x + C

Solution steps

\int \frac{x ^{2} + 9}{x ^{2} + 4} d x

\frac{x ^{2} + 9}{x ^{2} + 4} = \frac{5}{x ^{2} + 4} + 1

= \int \frac{5}{x ^{2} + 4} + 1 d x

Apply Integral Substitution

= \int 2 (\frac{5}{4 u ^{2} + 4} + 1) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{5}{4 u ^{2} + 4} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{5}{4 u ^{2} + 4} d u + \int 1 d u)

\int \frac{5}{4 u ^{2} + 4} d u = \frac{5}{4} arctan (u)

\int 1 d u = u

= 2 (\frac{5}{4} arctan (u) + u)

Substitute back

u = \frac{x}{2}

= 2 (\frac{5}{4} arctan (\frac{x}{2}) + \frac{x}{2})

Simplify

2 (\frac{5}{4} arctan (\frac{x}{2}) + \frac{x}{2}) : \frac{5}{2} arctan (\frac{x}{2}) + x

= \frac{5}{2} arctan (\frac{x}{2}) + x

Add a constant to the solution

= \frac{5}{2} arctan (\frac{x}{2}) + x + C

\int \frac{1}{7 y} d y

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\frac{d}{d x} ((arctanh (3 x^{2}))^{3})

\int x sin (z) d y

f (x) = (x^{3} - 16 x)^{8}, at x = 4

What is the integral of (x^2+9)/(x^2+4) ?
The integral of (x^2+9)/(x^2+4) is 5/2 arctan(x/2)+x+C