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

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

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

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

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

Solution steps

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

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 + r^{2}

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

Add a constant to the solution

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

\int_{0}^{4} x 4 - x d x

x \to \infty lim (\frac{( x + 1 ) ^{2} e ^{x^{3}}}{e ^{(x + 1)^{3}} x ^{2}})

x \to 16 - lim (\frac{8}{11 + 2 ^{\frac{1}{x - 16}}})

\int 2 (x^{2} + 2 x)^{16} (2 x + 2) d x

\frac{2 s}{( s ^{2} + 2 s + 6 )}

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