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

The integral of x/(sqrt(x+10)) is 2(1/3 (x+10)^{3/2}-10sqrt(x+10))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

2 (\frac{1}{3} (x + 10)^{\frac{3}{2}} - 10 x + 10) + C

Solution steps

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

Apply u-substitution

: \int 2 (u^{2} - 10) d u

= \int 2 (u^{2} - 10) 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} - 10 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 u^{2} d u - \int 10 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 10 d u = 10 u

= 2 (\frac{u ^{3}}{3} - 10 u)

Substitute back

u = x + 10

= 2 (\frac{( x + 10 ) ^{3}}{3} - 10 x + 10)

Simplify

2 (\frac{( x + 10 ) ^{3}}{3} - 10 x + 10) : 2 (\frac{1}{3} (x + 10)^{\frac{3}{2}} - 10 x + 10)

= 2 (\frac{1}{3} (x + 10)^{\frac{3}{2}} - 10 x + 10)

Add a constant to the solution

= 2 (\frac{1}{3} (x + 10)^{\frac{3}{2}} - 10 x + 10) + C

\int_{0}^{x} \frac{e ^{t^{2}} - 1}{t} d t

n = 1 \sum \infty \frac{n ^{5}}{3 ^{n}}

\int sec (x) (tan (x) + cos (x)) d x

3 y^{^{'}} = \frac{3 x \cdot y + 18 x ^{2} + 2 y ^{2}}{x ^{2}}, y (1) = 3

f (x) = x^{4} (3 - x)^{2}, at x = 1

What is the integral of x/(sqrt(x+10)) ?
The integral of x/(sqrt(x+10)) is 2(1/3 (x+10)^{3/2}-10sqrt(x+10))+C