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

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

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integral of sqrt(3x^2)

\int 3 x^{2} d x

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

Solution steps

\int 3 x^{2} d x

Apply radical rule:

n ab = n a n b,

assuming

a \geq 0, b \geq 0

= \int 3 x^{2} d x

Take the constant out:

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

= 3 \cdot \int x^{2} d x

x^{2} = (x),

assuming

x \geq 0

= 3 \cdot \int x d x

Apply the Power Rule

: \frac{x ^{2}}{2}

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Add a constant to the solution

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

f (t) = t

2 u

\int \frac{1}{7 x + 5} d x

y^{^{''}} + 9 y = sec (3 t)

n = 0 \sum \infty 2 (\frac{4}{5})^{n}

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