What is the integral of (5-x)/((x-4)(x-pi)) ?

The integral of (5-x)/((x-4)(x-pi)) is 1/(-pi+4)ln|x-4|+(5-pi)/(pi-4)ln|x-pi|+C

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integral of (5-x)/((x-4)(x-pi))

\int \frac{5 - x}{( x - 4 ) ( x - π )} d x

\frac{1}{- π + 4} ln ∣ x - 4 ∣ + \frac{5 - π}{π - 4} ln ∣ x - π ∣ + C

Solution steps

\int \frac{5 - x}{( x - 4 ) ( x - π )} d x

Take the partial fraction of

\frac{5 - x}{( x - 4 ) ( x - π )} : \frac{1}{( - π + 4 ) ( x - 4 )} + \frac{5 - π}{( π - 4 ) ( x - π )}

= \int \frac{1}{( - π + 4 ) ( x - 4 )} + \frac{5 - π}{( π - 4 ) ( x - π )} 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}{( - π + 4 ) ( x - 4 )} d x + \int \frac{5 - π}{( π - 4 ) ( x - π )} d x

\int \frac{1}{( - π + 4 ) ( x - 4 )} d x = \frac{1}{- π + 4} ln ∣ x - 4 ∣

\int \frac{5 - π}{( π - 4 ) ( x - π )} d x = \frac{5 - π}{π - 4} ln ∣ x - π ∣

= \frac{1}{- π + 4} ln ∣ x - 4 ∣ + \frac{5 - π}{π - 4} ln ∣ x - π ∣

Add a constant to the solution

= \frac{1}{- π + 4} ln ∣ x - 4 ∣ + \frac{5 - π}{π - 4} ln ∣ x - π ∣ + C

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

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

\int \frac{1}{7 x + 1} d x

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

\int \frac{1}{( 5 + 4 x ) ^{2}} d x

What is the integral of (5-x)/((x-4)(x-pi)) ?
The integral of (5-x)/((x-4)(x-pi)) is 1/(-pi+4)ln|x-4|+(5-pi)/(pi-4)ln|x-pi|+C