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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

\int sin^{2} (2 x + 5) d x

\frac{\partial}{\partial x} (e^{- 7 y} cos (7 x))

\frac{d y}{d x} = \frac{y}{x} + \frac{x}{2 ( y + x )}

\int_{0}^{1} (2 x + 1)^{3} d x

f (x) = 3 x + 7, (3, 4)

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