What is the integral of (3x^2+1)/(x^2+4) ?

The integral of (3x^2+1)/(x^2+4) is 3x-11/2 arctan(x/2)+C

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integral of (3x^2+1)/(x^2+4)

\int \frac{3 x ^{2} + 1}{x ^{2} + 4} d x

3 x - \frac{11}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{3 x ^{2} + 1}{x ^{2} + 4} d x

Expand

\frac{3 x ^{2} + 1}{x ^{2} + 4} : \frac{3 x ^{2}}{x ^{2} + 4} + \frac{1}{x ^{2} + 4}

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

\int \frac{3 x ^{2}}{x ^{2} + 4} d x = 6 (- arctan (\frac{x}{2}) + \frac{x}{2})

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

= 6 (- arctan (\frac{x}{2}) + \frac{x}{2}) + \frac{1}{2} arctan (\frac{x}{2})

Simplify

6 (- arctan (\frac{x}{2}) + \frac{x}{2}) + \frac{1}{2} arctan (\frac{x}{2}) : 3 x - \frac{11}{2} arctan (\frac{x}{2})

= 3 x - \frac{11}{2} arctan (\frac{x}{2})

Add a constant to the solution

= 3 x - \frac{11}{2} arctan (\frac{x}{2}) + C

\int 3 x (x + 4) d x

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

\int \frac{t}{t} d t

\int \frac{x ^{2} - x + 1}{x + 1} d x

in t e g r a l \frac{1}{2 ^{x}}

What is the integral of (3x^2+1)/(x^2+4) ?
The integral of (3x^2+1)/(x^2+4) is 3x-11/2 arctan(x/2)+C