What is the integral of sqrt(5x^2-4) ?

The integral of sqrt(5x^2-4) is 1/2 xsqrt(5x^2-4)-2/(sqrt(5))ln((|sqrt(5)x+sqrt(5x^2-4)|)/2)+C

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integral of sqrt(5x^2-4)

\int 5 x^{2} - 4 d x

\frac{1}{2} x 5 x^{2} - 4 - \frac{2}{5} ln (\frac{5 x + 5 x ^{2} - 4}{2}) + C

Solution steps

\int 5 x^{2} - 4 d x

Apply Trigonometric Substitution

: \int \frac{4}{5} tan^{2} (u) sec (u) d u

= \int \frac{4}{5} tan^{2} (u) sec (u) d u

Take the constant out:

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

= \frac{4}{5} \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= \frac{4}{5} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= \frac{4}{5} \cdot \int - sec (u) + sec^{3} (u) d u

Apply the Sum Rule:

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

= \frac{4}{5} (- \int sec (u) d u + \int sec^{3} (u) d u)

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

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= \frac{4}{5} (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{5}{2} x)

= \frac{4}{5} (- ln tan (arcsec (\frac{5}{2} x)) + sec (arcsec (\frac{5}{2} x)) + \frac{1}{2} sec (arcsec (\frac{5}{2} x)) tan (arcsec (\frac{5}{2} x)) + \frac{1}{2} ln tan (arcsec (\frac{5}{2} x)) + sec (arcsec (\frac{5}{2} x)))

Simplify

\frac{4}{5} (- ln tan (arcsec (\frac{5}{2} x)) + sec (arcsec (\frac{5}{2} x)) + \frac{1}{2} sec (arcsec (\frac{5}{2} x)) tan (arcsec (\frac{5}{2} x)) + \frac{1}{2} ln tan (arcsec (\frac{5}{2} x)) + sec (arcsec (\frac{5}{2} x))) : \frac{1}{2} x 5 x^{2} - 4 - \frac{2}{5} ln (\frac{5 x + 5 x ^{2} - 4}{2})

= \frac{1}{2} x 5 x^{2} - 4 - \frac{2}{5} ln (\frac{5 x + 5 x ^{2} - 4}{2})

Add a constant to the solution

= \frac{1}{2} x 5 x^{2} - 4 - \frac{2}{5} ln (\frac{5 x + 5 x ^{2} - 4}{2}) + C

\frac{d ^{2}}{d x ^{2}} (e^{\frac{- x ^{2}}{2}})

y = 7 x - 6 x, (1, 1)

y^{^{''}} + 9 y^{^{'}} - 10 y = 0

\frac{d y}{d x} = 2 + y - 2 x + 3

\int \frac{x ^{3}}{( x ^{2} + 16 ) ^{3}} d x

What is the integral of sqrt(5x^2-4) ?
The integral of sqrt(5x^2-4) is 1/2 xsqrt(5x^2-4)-2/(sqrt(5))ln((|sqrt(5)x+sqrt(5x^2-4)|)/2)+C