What is the integral of sqrt(1+3x^2)(6x) ?

The integral of sqrt(1+3x^2)(6x) is 2/3 (1+3x^2)^{3/2}+C

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integral of sqrt(1+3x^2)(6x)

\int 1 + 3 x^{2} (6 x) d x

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

Solution steps

\int 1 + 3 x^{2} \cdot 6 x d x

Take the constant out:

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

= 6 \cdot \int 1 + 3 x^{2} x d x

Apply u-substitution

: \int \frac{u}{6} d u

= 6 \cdot \int \frac{u}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 6 \cdot \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 3 x^{2}

= 6 \cdot \frac{1}{6} \cdot \frac{2}{3} (1 + 3 x^{2})^{\frac{3}{2}}

Simplify

6 \cdot \frac{1}{6} \cdot \frac{2}{3} (1 + 3 x^{2})^{\frac{3}{2}} : \frac{2}{3} (1 + 3 x^{2})^{\frac{3}{2}}

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

Add a constant to the solution

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

\int cos (\frac{x}{3}) d x

\int \frac{2 x ^{3} + 2 x ^{2} + 6 x + 9}{x ^{3} + 3 x} d x

\int (x^{2} - 8)^{5} (2 x) d x

\frac{\partial}{\partial y} (x^{2} y + \frac{x e ^{x}}{y ^{3}})

2 x^{2} - 2 x - 6, - 2, 2

What is the integral of sqrt(1+3x^2)(6x) ?
The integral of sqrt(1+3x^2)(6x) is 2/3 (1+3x^2)^{3/2}+C