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

The integral of (sqrt(x-3))/(x+4) is 2sqrt(7)(-arctan(sqrt(1/7 (x-3)))+sqrt(1/7 (x-3)))+C

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integral of (sqrt(x-3))/(x+4)

\int \frac{x - 3}{x + 4} d x

27 (- arctan (\frac{1}{7} (x - 3)) + \frac{1}{7} (x - 3)) + C

Solution steps

\int \frac{x - 3}{x + 4} d x

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} + 7} d u

= \int \frac{2 u ^{2}}{u ^{2} + 7} 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}}{u ^{2} + 7} d u

\frac{u ^{2}}{u ^{2} + 7} = - \frac{7}{u ^{2} + 7} + 1

= 2 \cdot \int - \frac{7}{u ^{2} + 7} + 1 d u

Apply Integral Substitution

= 2 \cdot \int 7 (- \frac{1}{v ^{2} + 1} + 1) d v

Take the constant out:

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

= 27 \cdot \int - \frac{1}{v ^{2} + 1} + 1 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

= 27 (- \int \frac{1}{v ^{2} + 1} d v + \int 1 d v)

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

\int 1 d v = v

= 27 (- arctan (v) + v)

Substitute back

= 27 (- arctan (\frac{x - 3}{7}) + \frac{x - 3}{7})

Simplify

27 (- arctan (\frac{x - 3}{7}) + \frac{x - 3}{7}) : 27 (- arctan (\frac{1}{7} (x - 3)) + \frac{1}{7} (x - 3))

= 27 (- arctan (\frac{1}{7} (x - 3)) + \frac{1}{7} (x - 3))

Add a constant to the solution

= 27 (- arctan (\frac{1}{7} (x - 3)) + \frac{1}{7} (x - 3)) + C

\int - 16 x d x

\int (x + 1)^{10} (x + 2) d x

\frac{1}{x - 6}

x, \frac{1}{x}, \frac{1}{2}

y = ln (sec (x))

What is the integral of (sqrt(x-3))/(x+4) ?
The integral of (sqrt(x-3))/(x+4) is 2sqrt(7)(-arctan(sqrt(1/7 (x-3)))+sqrt(1/7 (x-3)))+C