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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u + 2}{2 u} d u

= \int \frac{u + 2}{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 + 2}{u} d u

Expand

\frac{u + 2}{u} : 1 + \frac{2}{u}

= \frac{1}{2} \cdot \int 1 + \frac{2}{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{2}{u} d u)

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= \frac{1}{2} (u + 2 ln ∣ u ∣)

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{( ln ( x ^{6} ) )}{x} d x

f (x) = (2 - 3 x^{3})^{- 2}

\int_{0}^{π} 9 sin^{4} (3 t) d t

6 - 2 e^{x}

x \to 0 lim (2 cos (\frac{π}{x}))

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