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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

\int \frac{5}{( x - 1 ) ^{2}} d x = - \frac{5}{x - 1}

= 3 ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{5}{x - 1}

Add a constant to the solution

= 3 ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{5}{x - 1} + C

\frac{1}{( x ^{2} - 2 x - 5 ) ^{4}}

\int_{0}^{2} \frac{1}{2} y^{2} d y

sin (\frac{t}{t + 9})

\int \frac{2 x + 1}{3 x ^{2} - 27} d x

\int \frac{1}{e ^{- x} + 1} d x

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