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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

2 (- arctan (3 - x) + 3 - x) + C

Solution steps

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

Apply u-substitution

: \int - \frac{2 u ^{2}}{- u ^{2} - 1} d u

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

Factor

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

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

Take the constant out:

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

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

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

= - 2 (- \int - \frac{1}{u ^{2} + 1} + 1 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

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

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

\int 1 d u = u

= - 2 (- (- arctan (u) + u))

Substitute back

u = 3 - x

= - 2 (- (- arctan (3 - x) + 3 - x))

Simplify

= 2 (- arctan (3 - x) + 3 - x)

Add a constant to the solution

= 2 (- arctan (3 - x) + 3 - x) + C

(0, 2), (4, - 1)

\int \frac{sin ^{7} ( 2 x )}{cos ^{4} ( 2 x )} d x

\frac{\partial}{\partial x} (x y^{3} z^{2})

f (x) = x^{3} - 9 x + 5, at x = 3

y = sec (x^{2} + 3 x)

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