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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

- (x)^{2} 1

\frac{d}{d t} (- e^{3 t})

\int 6 x^{2} cos (x) d x

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

\frac{\partial}{\partial y} (2 e^{2 z} x y - 2 y)

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