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

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

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

\int \frac{( x ^{2} + x - 10 )}{( 2 x - 3 ) ( x ^{2} + 4 )} d x

- \frac{1}{2} ln ∣ 2 x - 3 ∣ + \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + arctan (\frac{x}{2}) + C

Solution steps

\int \frac{x ^{2} + x - 10}{( 2 x - 3 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{x ^{2} + x - 10}{( 2 x - 3 ) ( x ^{2} + 4 )} : - \frac{1}{2 x - 3} + \frac{x + 2}{x ^{2} + 4}

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

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

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

= - \frac{1}{2} ln ∣ 2 x - 3 ∣ + \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + arctan (\frac{x}{2})

Add a constant to the solution

= - \frac{1}{2} ln ∣ 2 x - 3 ∣ + \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + arctan (\frac{x}{2}) + C

\frac{d}{d x} (- 6 sin (3 x))

\int \frac{1}{cos ( x ) + 2 sin ( x ) + 3} d x

n = 2 \sum \infty \frac{ln ( n ^{2} )}{n}

\frac{( s ^{2} + 1 )}{s + 5}

\int_{0}^{4} (2 x + 4)^{- \frac{1}{2}} d x

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