What is the integral of sqrt((6t)^2+(6t^2)^2) ?

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

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integral of sqrt((6t)^2+(6t^2)^2)

\int (6 t)^{2} + (6 t^{2})^{2} d t

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

Solution steps

\int (6 t)^{2} + (6 t^{2})^{2} d t

Simplify

(6 t)^{2} + (6 t^{2})^{2} : 6 t^{2} 1 + t^{2}

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

Take the constant out:

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

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

t^{2} = (t),

assuming

t \geq 0

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

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 + t^{2}

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

Simplify

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

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

Add a constant to the solution

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

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