What is the integral of 2/(t^2)sqrt(1/t+9) ?

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

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integral of 2/(t^2)sqrt(1/t+9)

\int \frac{2}{t ^{2}} \frac{1}{t} + 9 d t

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

Solution steps

\int \frac{2}{t ^{2}} \frac{1}{t} + 9 d t

Take the constant out:

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

= 2 \cdot \int \frac{1}{t ^{2}} \frac{1}{t} + 9 d t

Apply u-substitution

: \int - u + 9 d u

= 2 \cdot \int - u + 9 d u

Take the constant out:

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

= 2 (- \int u + 9 d u)

Apply u-substitution

: \int v d v

= 2 (- \int v d v)

Apply the Power Rule

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

= 2 (- \frac{2}{3} v^{\frac{3}{2}})

Substitute back

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

Simplify

2 (- \frac{2}{3} (\frac{1}{t} + 9)^{\frac{3}{2}}) : - \frac{4}{3} (\frac{1 + 9 t}{t})^{\frac{3}{2}}

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

Add a constant to the solution

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

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