What is the integral of (x^2+19x+10)/(2x^4+5x^3) ?

The integral of (x^2+19x+10)/(2x^4+5x^3) is -ln|x|-3/x-1/(x^2)+ln|2x+5|+C

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integral of (x^2+19x+10)/(2x^4+5x^3)

\int \frac{x ^{2} + 19 x + 10}{2 x ^{4} + 5 x ^{3}} d x

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

Solution steps

\int \frac{x ^{2} + 19 x + 10}{2 x ^{4} + 5 x ^{3}} d x

Take the partial fraction of

\frac{x ^{2} + 19 x + 10}{2 x ^{4} + 5 x ^{3}} : - \frac{1}{x} + \frac{3}{x ^{2}} + \frac{2}{x ^{3}} + \frac{2}{2 x + 5}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{2}{x ^{3}} d x = - \frac{1}{x ^{2}}

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

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

Add a constant to the solution

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

What is the integral of (x^2+19x+10)/(2x^4+5x^3) ?
The integral of (x^2+19x+10)/(2x^4+5x^3) is -ln|x|-3/x-1/(x^2)+ln|2x+5|+C