What is the integral of sqrt(4x^2+144x^4) ?

The integral of sqrt(4x^2+144x^4) is 1/54 (1+36x^2)^{3/2}+C

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integral of sqrt(4x^2+144x^4)

\int 4 x^{2} + 144 x^{4} d x

\frac{1}{54} (1 + 36 x^{2})^{\frac{3}{2}} + C

Solution steps

\int 4 x^{2} + 144 x^{4} d x

Factor

4 x^{2} + 144 x^{4} : 2 x^{2} 1 + 36 x^{2}

= \int 2 x^{2} 1 + 36 x^{2} d x

Take the constant out:

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

= 2 \cdot \int x^{2} 1 + 36 x^{2} d x

x^{2} = (x),

assuming

x \geq 0

= 2 \cdot \int x 1 + 36 x^{2} d x

Apply u-substitution

: \int \frac{u}{72} d u

= 2 \cdot \int \frac{u}{72} d u

Take the constant out:

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

= 2 \cdot \frac{1}{72} \cdot \int u d u

Apply the Power Rule

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

= 2 \cdot \frac{1}{72} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 36 x^{2}

= 2 \cdot \frac{1}{72} \cdot \frac{2}{3} (1 + 36 x^{2})^{\frac{3}{2}}

Simplify

2 \cdot \frac{1}{72} \cdot \frac{2}{3} (1 + 36 x^{2})^{\frac{3}{2}} : \frac{1}{54} (1 + 36 x^{2})^{\frac{3}{2}}

= \frac{1}{54} (1 + 36 x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{54} (1 + 36 x^{2})^{\frac{3}{2}} + C

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What is the integral of sqrt(4x^2+144x^4) ?
The integral of sqrt(4x^2+144x^4) is 1/54 (1+36x^2)^{3/2}+C