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

The integral of sqrt((3t^2)^2+(2t)^2) is 1/27 (9t^2+4)^{3/2}+C

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integral of sqrt((3t^2)^2+(2t)^2)

\int (3 t^{2})^{2} + (2 t)^{2} d t

\frac{1}{27} (9 t^{2} + 4)^{\frac{3}{2}} + C

Solution steps

\int (3 t^{2})^{2} + (2 t)^{2} d t

Factor

(3 t^{2})^{2} + (2 t)^{2} : t^{2} 9 t^{2} + 4

= \int t^{2} 9 t^{2} + 4 d t

t^{2} = (t),

assuming

t \geq 0

= \int t 9 t^{2} + 4 d t

Apply u-substitution

: \int \frac{u}{18} d u

= \int \frac{u}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{18} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 9 t^{2} + 4

= \frac{1}{18} \cdot \frac{2}{3} (9 t^{2} + 4)^{\frac{3}{2}}

Simplify

\frac{1}{18} \cdot \frac{2}{3} (9 t^{2} + 4)^{\frac{3}{2}} : \frac{1}{27} (9 t^{2} + 4)^{\frac{3}{2}}

= \frac{1}{27} (9 t^{2} + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{27} (9 t^{2} + 4)^{\frac{3}{2}} + C

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What is the integral of sqrt((3t^2)^2+(2t)^2) ?
The integral of sqrt((3t^2)^2+(2t)^2) is 1/27 (9t^2+4)^{3/2}+C