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

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

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

\int \frac{3}{x ^{4} + 2 x ^{3} + x ^{2}} d x

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

Solution steps

\int \frac{3}{x ^{4} + 2 x ^{3} + x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{4} + 2 x ^{3} + x ^{2}} d x

Take the partial fraction of

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

= 3 \cdot \int - \frac{2}{x} + \frac{1}{x ^{2}} + \frac{2}{x + 1} + \frac{1}{( 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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

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

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

Add a constant to the solution

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

\int 34 cos (x) cos^{3} (sin (x)) d x

f (x) = \frac{π x}{4}

x \to - \infty lim (2^{3 x})

\frac{d v}{d t} = - 2 v, v (0) = 4

\frac{\partial}{\partial x} (3 x^{2} - 4 x y)

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