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

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

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integral of tsqrt(4+9t^2)

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{1 + 64 x ^{2}} d x

\int_{0}^{\infty} t e^{- t^{2}} d t

\frac{d}{d x} (\frac{x}{x - 9})

\int (3 x - 5)^{\frac{1}{2}} d x

y^{^{''}} = 6 x - 12

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