What is the integral of 1/(ysqrt(5y^2-3)) ?

The integral of 1/(ysqrt(5y^2-3)) is 1/(sqrt(3))arcsec(sqrt(5/3)y)+C

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integral of 1/(ysqrt(5y^2-3))

\int \frac{1}{y 5 y ^{2} - 3} d y

\frac{1}{3} arcsec (\frac{5}{3} y) + C

Solution steps

\int \frac{1}{y 5 y ^{2} - 3} d y

Apply Trigonometric Substitution

: \int \frac{1}{3} d u

= \int \frac{1}{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 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{5}{3} y)

= \frac{1}{3} \cdot 1 \cdot arcsec (\frac{5}{3} y)

Simplify

\frac{1}{3} \cdot 1 \cdot arcsec (\frac{5}{3} y) : \frac{1}{3} arcsec (\frac{5}{3} y)

= \frac{1}{3} arcsec (\frac{5}{3} y)

Add a constant to the solution

= \frac{1}{3} arcsec (\frac{5}{3} y) + C

\int (x - 0.1)^{2} (x - 0.75)^{2} d x

\int \frac{ln ( x )}{4} d x

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

\frac{\partial}{\partial x} (y^{0.75} x^{0.25})

f (x) = 9 e^{5} x + e^{- x^{2}}, at x = 1

What is the integral of 1/(ysqrt(5y^2-3)) ?
The integral of 1/(ysqrt(5y^2-3)) is 1/(sqrt(3))arcsec(sqrt(5/3)y)+C