What is the integral of tsqrt(9t^2+4) ?

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

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integral of tsqrt(9t^2+4)

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

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

Solution steps

\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

x^{2}, x^{4} - 3 x^{2}, - 2, 2

x \to a lim (x^{s i n (x)})

\frac{x ^{2}}{2 x + 4}

\frac{d x}{d t} - 2 x = x^{3}

\frac{d}{d x} (ln (x) - 2 x^{2})

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