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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

\frac{d}{d x} (\frac{7}{e ^{x}})

\int (1 + \frac{1}{x})^{- 2} (\frac{1}{x ^{2}}) d x

y = 9 x^{3} - x^{2} + 5 x - 6

\int e^{0.02 x} d x

\frac{d}{d x} (4 x ln (x))

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