What is the integral of (sqrt(2y^2-3))/(y^2) ?

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

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integral of (sqrt(2y^2-3))/(y^2)

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

2 ln \frac{1}{3} (2 y^{2} - 3) + \frac{2}{3} y - \frac{2 y ^{2} - 3}{y} + C

Solution steps

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

Apply Integral Substitution

= \int \frac{2 u ^{2} - 6}{u ^{2}} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2} - 6}{u ^{2}} d u

Apply Trigonometric Substitution

: \int tan^{2} (v) cos (v) d v

= 2 \cdot \int tan^{2} (v) cos (v) d v

Simplify

tan^{2} (v) cos (v) : sin (v) tan (v)

= 2 \cdot \int sin (v) tan (v) d v

Rewrite using trig identities

= 2 \cdot \int \frac{1 - cos ^{2} ( v )}{cos ( v )} d v

Expand

\frac{1 - cos ^{2} ( v )}{cos ( v )} : \frac{1}{cos ( v )} - cos (v)

= 2 \cdot \int \frac{1}{cos ( v )} - cos (v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{1}{cos ( v )} d v - \int cos (v) d v)

\int \frac{1}{cos ( v )} d v = ln ∣ tan (v) + sec (v) ∣

\int cos (v) d v = sin (v)

= 2 (ln ∣ tan (v) + sec (v) ∣ - sin (v))

Substitute back

= 2 (ln tan (arcsec (\frac{1}{6} \cdot 2 y)) + sec (arcsec (\frac{1}{6} \cdot 2 y)) - sin (arcsec (\frac{1}{6} \cdot 2 y)))

Simplify

2 (ln tan (arcsec (\frac{1}{6} \cdot 2 y)) + sec (arcsec (\frac{1}{6} \cdot 2 y)) - sin (arcsec (\frac{1}{6} \cdot 2 y))) : 2 ln \frac{1}{3} (2 y^{2} - 3) + \frac{2}{3} y - \frac{2 y ^{2} - 3}{y}

= 2 ln \frac{1}{3} (2 y^{2} - 3) + \frac{2}{3} y - \frac{2 y ^{2} - 3}{y}

Add a constant to the solution

= 2 ln \frac{1}{3} (2 y^{2} - 3) + \frac{2}{3} y - \frac{2 y ^{2} - 3}{y} + C

\frac{d}{d x} ((2 x)^{4 x})

\int (x^{2} + 5) d x

\frac{cos ( ln ( x ) )}{x}

\int \frac{x + 1}{x} d x

y^{^{'}} = 4 y^{2} + 1

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