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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int sec (u) d u

= 2 \cdot \int sec (u) d u

Use the common integral:

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

= 2 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{1}{6} x)

= 2 ln tan (arctan (\frac{1}{6} x)) + sec (arctan (\frac{1}{6} x))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (sin (x) (- 2 x))

x \to 4 lim (1)

\frac{d}{d x} (\frac{x ^{\frac{6}{7}}}{5})

\frac{1}{x ^{2} + 3 x - 1}

\int \frac{1}{( x - 1 ) ^{2} + 4} d x

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