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

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

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

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

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

Solution steps

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

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} t + 1)) + sec (arcsec (\frac{1}{2} t + 1)) + \frac{1}{2} sec (arcsec (\frac{1}{2} t + 1)) tan (arcsec (\frac{1}{2} t + 1)) + \frac{1}{2} ln tan (arcsec (\frac{1}{2} t + 1)) + sec (arcsec (\frac{1}{2} t + 1)))

Simplify

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

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

Add a constant to the solution

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

y = x^{3} - 2 x^{2} + 2, y = 3 x^{2} + 5 x - 23

\frac{( x ^{2} + 1 ) d y}{d x} + 6 x (y - 1) = 0, y (0) = 3

t \frac{d y}{d t} = (t^{3} + 22 t^{3} \cdot y)

\int \frac{x ^{8}}{1 + x ^{18}} d x

\frac{d}{d x} (2 x e^{x^{2} - 1})

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