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

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

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

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

\frac{2}{3} x^{3} + 3 x + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{3} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{3} + 3 x

= \frac{1}{3} \cdot 2 (x^{3} + 3 x)^{\frac{1}{2}}

Simplify

\frac{1}{3} \cdot 2 (x^{3} + 3 x)^{\frac{1}{2}} : \frac{2}{3} x^{3} + 3 x

= \frac{2}{3} x^{3} + 3 x

Add a constant to the solution

= \frac{2}{3} x^{3} + 3 x + C

\int e^{x} \cdot (- 3) e^{e^{x} + 2} d x

\frac{d}{d x} (x 5^{2 x})

\int_{0}^{3} 1

\int \frac{9 + x ^{2}}{x ^{4}} d x

y = x^{3} - 9 x, (1, - 8)

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