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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

x (1 - 9 x^{3})

x \to 2 + lim (2 x + 2 + x)

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

\int e^{- 0.6 x} d x

y = 9 x^{2} + 3 x - 9

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