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

The integral of (x^2+x+1)/((x-1)^2(x+3)) is 9/16 ln|x-1|-3/(4(x-1))+7/16 ln|x+3|+C

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

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

\frac{9}{16} ln ∣ x - 1 ∣ - \frac{3}{4 ( x - 1 )} + \frac{7}{16} ln ∣ x + 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} + x + 1}{( x - 1 ) ^{2} ( x + 3 )} : \frac{9}{16 ( x - 1 )} + \frac{3}{4 ( x - 1 ) ^{2}} + \frac{7}{16 ( x + 3 )}

= \int \frac{9}{16 ( x - 1 )} + \frac{3}{4 ( x - 1 ) ^{2}} + \frac{7}{16 ( 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{9}{16 ( x - 1 )} d x + \int \frac{3}{4 ( x - 1 ) ^{2}} d x + \int \frac{7}{16 ( x + 3 )} d x

\int \frac{9}{16 ( x - 1 )} d x = \frac{9}{16} ln ∣ x - 1 ∣

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

\int \frac{7}{16 ( x + 3 )} d x = \frac{7}{16} ln ∣ x + 3 ∣

= \frac{9}{16} ln ∣ x - 1 ∣ - \frac{3}{4 ( x - 1 )} + \frac{7}{16} ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{9}{16} ln ∣ x - 1 ∣ - \frac{3}{4 ( x - 1 )} + \frac{7}{16} ln ∣ x + 3 ∣ + C

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What is the integral of (x^2+x+1)/((x-1)^2(x+3)) ?
The integral of (x^2+x+1)/((x-1)^2(x+3)) is 9/16 ln|x-1|-3/(4(x-1))+7/16 ln|x+3|+C