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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{2 u} d u

= \int \frac{u - 1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \frac{1}{2} \cdot \int 1 - \frac{1}{u} d u

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 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{2} (u - ln ∣ u ∣)

Substitute back

u = 1 + x^{2}

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

Add a constant to the solution

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

x^{- \frac{1}{4}} + x^{\frac{1}{4}}

f (x) = x^{2} - 7, at x = 3

\frac{2}{x}, \frac{2}{x ^{2}}, 4

\frac{d}{d x} (4 arctan (x))

(y - x) e^{x} + (1 + e^{x}) y^{^{'}} = 0, y (1) = 1

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