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

The integral of (2x+3)/((x-1)^4(x-2)) is -7ln|x-1|+7/(x-1)+7/(2(x-1)^2)+5/(3(x-1)^3)+7ln|x-2|+C

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integral of (2x+3)/((x-1)^4(x-2))

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

- 7 ln ∣ x - 1 ∣ + \frac{7}{x - 1} + \frac{7}{2 ( x - 1 ) ^{2}} + \frac{5}{3 ( x - 1 ) ^{3}} + 7 ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{2 x + 3}{( x - 1 ) ^{4} ( x - 2 )} : - \frac{7}{x - 1} - \frac{7}{( x - 1 ) ^{2}} - \frac{7}{( x - 1 ) ^{3}} - \frac{5}{( x - 1 ) ^{4}} + \frac{7}{x - 2}

= \int - \frac{7}{x - 1} - \frac{7}{( x - 1 ) ^{2}} - \frac{7}{( x - 1 ) ^{3}} - \frac{5}{( x - 1 ) ^{4}} + \frac{7}{x - 2} 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{7}{x - 1} d x - \int \frac{7}{( x - 1 ) ^{2}} d x - \int \frac{7}{( x - 1 ) ^{3}} d x - \int \frac{5}{( x - 1 ) ^{4}} d x + \int \frac{7}{x - 2} d x

\int \frac{7}{x - 1} d x = 7 ln ∣ x - 1 ∣

\int \frac{7}{( x - 1 ) ^{2}} d x = - \frac{7}{x - 1}

\int \frac{7}{( x - 1 ) ^{3}} d x = - \frac{7}{2 ( x - 1 ) ^{2}}

\int \frac{5}{( x - 1 ) ^{4}} d x = - \frac{5}{3 ( x - 1 ) ^{3}}

\int \frac{7}{x - 2} d x = 7 ln ∣ x - 2 ∣

= - 7 ln ∣ x - 1 ∣ - (- \frac{7}{x - 1}) - (- \frac{7}{2 ( x - 1 ) ^{2}}) - (- \frac{5}{3 ( x - 1 ) ^{3}}) + 7 ln ∣ x - 2 ∣

Simplify

= - 7 ln ∣ x - 1 ∣ + \frac{7}{x - 1} + \frac{7}{2 ( x - 1 ) ^{2}} + \frac{5}{3 ( x - 1 ) ^{3}} + 7 ln ∣ x - 2 ∣

Add a constant to the solution

= - 7 ln ∣ x - 1 ∣ + \frac{7}{x - 1} + \frac{7}{2 ( x - 1 ) ^{2}} + \frac{5}{3 ( x - 1 ) ^{3}} + 7 ln ∣ x - 2 ∣ + C

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What is the integral of (2x+3)/((x-1)^4(x-2)) ?
The integral of (2x+3)/((x-1)^4(x-2)) is -7ln|x-1|+7/(x-1)+7/(2(x-1)^2)+5/(3(x-1)^3)+7ln|x-2|+C