What is the integral of (x^4+x-25)/(x^2+5) ?

The integral of (x^4+x-25)/(x^2+5) is (x^3)/3-5x+1/2 ln|x^2+5|+C

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integral of (x^4+x-25)/(x^2+5)

\int \frac{x ^{4} + x - 25}{x ^{2} + 5} d x

\frac{x ^{3}}{3} - 5 x + \frac{1}{2} ln x^{2} + 5 + C

Solution steps

\int \frac{x ^{4} + x - 25}{x ^{2} + 5} d x

Long division

\frac{x ^{4} + x - 25}{x ^{2} + 5} : x^{2} - 5 + \frac{x}{x ^{2} + 5}

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

\int x^{2} d x = \frac{x ^{3}}{3}

\int 5 d x = 5 x

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

= \frac{x ^{3}}{3} - 5 x + \frac{1}{2} ln x^{2} + 5

Add a constant to the solution

= \frac{x ^{3}}{3} - 5 x + \frac{1}{2} ln x^{2} + 5 + C

\int x^{2} (1 - x)^{\frac{1}{2}} d x

\frac{d y}{d x} + 6 y = 0

\int_{2}^{3} x^{5} d x

f (x) = x^{2} - 2 x + 1, g (x) = - x + 3

\int tan^{5} (\frac{x}{4}) d x

What is the integral of (x^4+x-25)/(x^2+5) ?
The integral of (x^4+x-25)/(x^2+5) is (x^3)/3-5x+1/2 ln|x^2+5|+C