What is the integral of (sqrt(x-3))/(sqrt(x+1)) ?

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

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integral of (sqrt(x-3))/(sqrt(x+1))

\int \frac{x - 3}{x + 1} d x

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

Solution steps

\int \frac{x - 3}{x + 1} d x

Apply u-substitution

: \int 2 u^{2} - 4 d u

= \int 2 u^{2} - 4 d u

Take the constant out:

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

= 2 \cdot \int u^{2} - 4 d u

Apply Trigonometric Substitution

: \int 4 tan^{2} (v) sec (v) d v

= 2 \cdot \int 4 tan^{2} (v) sec (v) d v

Take the constant out:

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

= 2 \cdot 4 \cdot \int tan^{2} (v) sec (v) d v

Rewrite using trig identities

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

Expand

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

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

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int \frac{9}{x ( x + 3 )} d x

\int ln (y) d y

\int 3 x^{2} - 24 x + 36 d x

\frac{d}{d x} (e^{x} cos (y z))

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