What is the integral of sqrt(x^2-3/4) ?

The integral of sqrt(x^2-3/4) is 1/8 (2xsqrt(4x^2-3)-3(ln|2x+sqrt(3)sqrt((4x^2-3)/3)|-(ln(3))/2))+C

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integral of sqrt(x^2-3/4)

\int x^{2} - \frac{3}{4} d x

\frac{1}{8} (2 x 4 x^{2} - 3 - 3 (ln 2 x + 3 \frac{4 x ^{2} - 3}{3} - \frac{ln ( 3 )}{2})) + C

Solution steps

\int x^{2} - \frac{3}{4} d x

Apply Trigonometric Substitution

: \int \frac{3 tan ^{2} ( u ) sec ( u )}{4} d u

= \int \frac{3 tan ^{2} ( u ) sec ( u )}{4} d u

Take the constant out:

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

= \frac{3}{4} \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= \frac{3}{4} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= \frac{3}{4} \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{3}{4} (- \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{3}{4} (- 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{1}{\frac{3}{2}} x)

= \frac{3}{4} (- ln tan (arcsec (\frac{1}{\frac{3}{2}} x)) + sec (arcsec (\frac{1}{\frac{3}{2}} x)) + \frac{1}{2} sec (arcsec (\frac{1}{\frac{3}{2}} x)) tan (arcsec (\frac{1}{\frac{3}{2}} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{\frac{3}{2}} x)) + sec (arcsec (\frac{1}{\frac{3}{2}} x)))

Simplify

\frac{3}{4} (- ln tan (arcsec (\frac{1}{\frac{3}{2}} x)) + sec (arcsec (\frac{1}{\frac{3}{2}} x)) + \frac{1}{2} sec (arcsec (\frac{1}{\frac{3}{2}} x)) tan (arcsec (\frac{1}{\frac{3}{2}} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{\frac{3}{2}} x)) + sec (arcsec (\frac{1}{\frac{3}{2}} x))) : \frac{1}{8} (2 x 4 x^{2} - 3 - 3 (ln 2 x + 3 \frac{4 x ^{2} - 3}{3} - \frac{ln ( 3 )}{2}))

= \frac{1}{8} (2 x 4 x^{2} - 3 - 3 (ln 2 x + 3 \frac{4 x ^{2} - 3}{3} - \frac{ln ( 3 )}{2}))

Add a constant to the solution

= \frac{1}{8} (2 x 4 x^{2} - 3 - 3 (ln 2 x + 3 \frac{4 x ^{2} - 3}{3} - \frac{ln ( 3 )}{2})) + C

\frac{d}{d x} (sin (π))

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

y^{^{''}} - 2 y = 5

\int \frac{( 3 in x - 5 ) ^{4}}{x} d x

\frac{d y}{d x} = 5 y^{2} - \frac{y}{x}

What is the integral of sqrt(x^2-3/4) ?
The integral of sqrt(x^2-3/4) is 1/8 (2xsqrt(4x^2-3)-3(ln|2x+sqrt(3)sqrt((4x^2-3)/3)|-(ln(3))/2))+C