What is the integral of (sqrt(x+1))/(x+10) ?

The integral of (sqrt(x+1))/(x+10) is 6(-arctan(1/3 sqrt(x+1))+1/3 sqrt(x+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(x+1))/(x+10)

\int \frac{x + 1}{x + 10} d x

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

Solution steps

\int \frac{x + 1}{x + 10} d x

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} + 9} d u

= \int \frac{2 u ^{2}}{u ^{2} + 9} 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} + 9} d u

\frac{u ^{2}}{u ^{2} + 9} = - \frac{9}{u ^{2} + 9} + 1

= 2 \cdot \int - \frac{9}{u ^{2} + 9} + 1 d u

Apply Integral Substitution

= 2 \cdot \int 3 (- \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

= 2 \cdot 3 \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

= 2 \cdot 3 (- \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

= 2 \cdot 3 (- arctan (v) + v)

Substitute back

= 2 \cdot 3 (- arctan (\frac{x + 1}{3}) + \frac{x + 1}{3})

Simplify

2 \cdot 3 (- arctan (\frac{x + 1}{3}) + \frac{x + 1}{3}) : 6 (- arctan (\frac{1}{3} x + 1) + \frac{1}{3} x + 1)

= 6 (- arctan (\frac{1}{3} x + 1) + \frac{1}{3} x + 1)

Add a constant to the solution

= 6 (- arctan (\frac{1}{3} x + 1) + \frac{1}{3} x + 1) + C

f (x) = x^{2}, (0, 3)

\int \frac{1}{e ^{x} e ^{2 x} - 9} d x

sin (x), 0, 2 π

\int x^{15} d x

\frac{d}{d x} ((7 x^{2} - 12 x) e^{x})

What is the integral of (sqrt(x+1))/(x+10) ?
The integral of (sqrt(x+1))/(x+10) is 6(-arctan(1/3 sqrt(x+1))+1/3 sqrt(x+1))+C