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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{4 u}{12 u + ( u - 1 ) ^{3} - 6 ( u - 1 ) ^{2} - 20} d u

= \int \frac{4 u}{12 u + ( u - 1 ) ^{3} - 6 ( u - 1 ) ^{2} - 20} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{u}{12 u + ( u - 1 ) ^{3} - 6 ( u - 1 ) ^{2} - 20} d u

Take the partial fraction of

\frac{u}{12 u + ( u - 1 ) ^{3} - 6 ( u - 1 ) ^{2} - 20} : \frac{1}{( u - 3 ) ^{2}} + \frac{3}{( u - 3 ) ^{3}}

= 4 \cdot \int \frac{1}{( u - 3 ) ^{2}} + \frac{3}{( u - 3 ) ^{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{3}{( u - 3 ) ^{3}} d u = - \frac{3}{2 ( u - 3 ) ^{2}}

= 4 (- \frac{1}{u - 3} - \frac{3}{2 ( u - 3 ) ^{2}})

Substitute back

u = 2 x + 1

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

Simplify

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

Add a constant to the solution

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

\frac{d y}{d x} x + y = \frac{1}{y ^{2}}

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

x \to - 1 - lim (∣ x - 1 ∣)

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

\frac{d}{d x} (x^{4} - 18 x^{2} + 1)

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