What is the integral of (30)/(sqrt(36+25x^2)) ?

The integral of (30)/(sqrt(36+25x^2)) is 6ln(1/6 |5x+sqrt(25x^2+36)|)+C

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integral of (30)/(sqrt(36+25x^2))

\int \frac{30}{36 + 25 x ^{2}} d x

6 ln (\frac{1}{6} 5 x + 25 x^{2} + 36) + C

Solution steps

\int \frac{30}{36 + 25 x ^{2}} d x

Take the constant out:

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

= 30 \cdot \int \frac{1}{36 + 25 x ^{2}} d x

Apply Trigonometric Substitution

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

= 30 \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

= 30 \cdot \frac{1}{5} \cdot \int sec (u) d u

Use the common integral:

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

= 30 \cdot \frac{1}{5} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{5}{6} x)

= 30 \cdot \frac{1}{5} ln tan (arctan (\frac{5}{6} x)) + sec (arctan (\frac{5}{6} x))

Simplify

30 \cdot \frac{1}{5} ln tan (arctan (\frac{5}{6} x)) + sec (arctan (\frac{5}{6} x)) : 6 ln (\frac{1}{6} 5 x + 25 x^{2} + 36)

= 6 ln (\frac{1}{6} 5 x + 25 x^{2} + 36)

Add a constant to the solution

= 6 ln (\frac{1}{6} 5 x + 25 x^{2} + 36) + C

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What is the integral of (30)/(sqrt(36+25x^2)) ?
The integral of (30)/(sqrt(36+25x^2)) is 6ln(1/6 |5x+sqrt(25x^2+36)|)+C