What is the integral of sqrt(36x^2+36x^4) ?

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

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integral of sqrt(36x^2+36x^4)

\int 36 x^{2} + 36 x^{4} d x

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

Solution steps

\int 36 x^{2} + 36 x^{4} d x

Simplify

36 x^{2} + 36 x^{4} : 6 x^{2} 1 + x^{2}

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

x^{2} = (x),

assuming

x \geq 0

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

Apply u-substitution

: \int u^{2} d u

= 6 \cdot \int u^{2} d u

Apply the Power Rule

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

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

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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