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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

\int \frac{3}{4 ( x + 1 )} d x = \frac{3}{4} ln ∣ x + 1 ∣

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

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

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

Simplify

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

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

Add a constant to the solution

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

f (x) = \frac{1}{5 - 2 e ^{x}}

\int 3 x - 6 d x

\int \frac{9 + 7 x}{1 + x ^{2}} d x

\int \frac{3 x ^{2} + 4}{0.3 x ^{3} + 1.2 x} d x

f (x) = x^{6} tan (x)

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