What is the integral of (6x^2+x+25)/((x+1)(x^2+4)) ?

The integral of (6x^2+x+25)/((x+1)(x^2+4)) is 6ln|x+1|+1/2 arctan(x/2)+C

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integral of (6x^2+x+25)/((x+1)(x^2+4))

\int \frac{6 x ^{2} + x + 25}{( x + 1 ) ( x ^{2} + 4 )} d x

6 ln ∣ x + 1 ∣ + \frac{1}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{6 x ^{2} + x + 25}{( x + 1 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{6 x ^{2} + x + 25}{( x + 1 ) ( x ^{2} + 4 )} : \frac{6}{x + 1} + \frac{1}{x ^{2} + 4}

= \int \frac{6}{x + 1} + \frac{1}{x ^{2} + 4} d x

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

\int \frac{6}{x + 1} d x = 6 ln ∣ x + 1 ∣

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

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

Add a constant to the solution

= 6 ln ∣ x + 1 ∣ + \frac{1}{2} arctan (\frac{x}{2}) + C

(x^{3} - 9) x

\frac{2 w ^{3} - w}{w}

\int t (t^{2} + 1)^{5} d t

n = 0 \sum \infty \frac{n - 1}{n ^{2}}

\int 1 + e^{t} d t

What is the integral of (6x^2+x+25)/((x+1)(x^2+4)) ?
The integral of (6x^2+x+25)/((x+1)(x^2+4)) is 6ln|x+1|+1/2 arctan(x/2)+C