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

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

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

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

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

Solution steps

\int \frac{1}{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

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

Take the partial fraction of

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Add a constant to the solution

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

y^{\frac{1}{2}} \frac{d y}{d x} + y^{\frac{3}{2}} = 1, y (0) = 9

17 e^{x} - e x^{e}

\frac{\partial}{\partial x} (e^{- x} sin (yz))

f (x) = 2 x^{5} - 6 x^{3}, at x = 1

\int ln (d) x d x

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