What is the integral of (-2)/(sqrt(81+25t^2)) ?

The integral of (-2)/(sqrt(81+25t^2)) is -2/5 ln((|5t+sqrt(81+25t^2)|)/9)+C

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integral of (-2)/(sqrt(81+25t^2))

\int \frac{- 2}{81 + 25 t ^{2}} d t

- \frac{2}{5} ln (\frac{5 t + 81 + 25 t ^{2}}{9}) + C

Solution steps

\int \frac{- 2}{81 + 25 t ^{2}} 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}{81 + 25 t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5} d u

= - 2 \cdot \int \frac{sec ( u )}{5} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{5} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= - 2 \cdot \frac{1}{5} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{5}{9} t)

= - 2 \cdot \frac{1}{5} ln tan (arctan (\frac{5}{9} t)) + sec (arctan (\frac{5}{9} t))

Simplify

- 2 \cdot \frac{1}{5} ln tan (arctan (\frac{5}{9} t)) + sec (arctan (\frac{5}{9} t)) : - \frac{2}{5} ln (\frac{5 t + 81 + 25 t ^{2}}{9})

= - \frac{2}{5} ln (\frac{5 t + 81 + 25 t ^{2}}{9})

Add a constant to the solution

= - \frac{2}{5} ln (\frac{5 t + 81 + 25 t ^{2}}{9}) + C

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What is the integral of (-2)/(sqrt(81+25t^2)) ?
The integral of (-2)/(sqrt(81+25t^2)) is -2/5 ln((|5t+sqrt(81+25t^2)|)/9)+C