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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{v - 1}{v} : 1 - \frac{1}{v}

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

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

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{1}{2} (v - ln ∣ v ∣)

Substitute back

v = 1 + u^{2}

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

Add a constant to the solution

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

\int π x d x

t \to + 0 lim (\frac{2 t}{tan ( t )})

x \to \infty lim (- x - 1)

\int \frac{1}{9 x ^{2} + 16} d x

h \to 0 + lim (\frac{∣ h ∣}{h} + h^{2})

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