What is the integral of 1/(sqrt(5+3x^2)) ?

The integral of 1/(sqrt(5+3x^2)) is 1/(sqrt(3))ln|sqrt(3/5)x+sqrt(1/5 (5+3x^2))|+C

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integral of 1/(sqrt(5+3x^2))

\int \frac{1}{5 + 3 x ^{2}} d x

\frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 3 x^{2}) + C

Solution steps

\int \frac{1}{5 + 3 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3} d u

= \int \frac{sec ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{3}{5} x)

= \frac{1}{3} ln tan (arctan (\frac{3}{5} x)) + sec (arctan (\frac{3}{5} x))

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{5} x)) + sec (arctan (\frac{3}{5} x)) : \frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 3 x^{2})

= \frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 3 x^{2})

Add a constant to the solution

= \frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 3 x^{2}) + C

24 x - 4 x

\frac{d y}{d x} = \frac{1}{2} x + 1

f (x) = \frac{1}{x}, g (x) = \frac{1}{3}, [2, 5]

\frac{7 u ^{2}}{( u ^{2} + u ) ^{3}}

\int \frac{3 x ^{2}}{1 - x ^{6}} d x

What is the integral of 1/(sqrt(5+3x^2)) ?
The integral of 1/(sqrt(5+3x^2)) is 1/(sqrt(3))ln|sqrt(3/5)x+sqrt(1/5 (5+3x^2))|+C