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

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

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

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

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

Solution steps

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

Expand

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

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

\int \frac{2 x ^{3}}{1 - x ^{2}} d x = 1 - x^{2} - ln 1 - x^{2}

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

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

Add a constant to the solution

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

\frac{d}{d x} (x^{4} - 8 x^{2} + 17)

\frac{d}{d x} (3 csc (x) + 6 cos (x))

\int \frac{x ^{4}}{4 - x ^{2}} d x

\frac{1}{5} x^{5}

\frac{\partial}{\partial p ( r , t )} (p (r, t) e^{r t})

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