What is the integral of (x^5)/(x^2sqrt(9x^2+9)) ?

The integral of (x^5)/(x^2sqrt(9x^2+9)) is 1/3 (1/3 (x^2+1)^{3/2}-sqrt(x^2+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^5)/(x^2sqrt(9x^2+9))

\int \frac{x ^{5}}{x ^{2} 9 x ^{2} + 9} d x

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

Solution steps

\int \frac{x ^{5}}{x ^{2} 9 x ^{2} + 9} d x

Simplify

\frac{x ^{5}}{x ^{2} 9 x ^{2} + 9} : \frac{x ^{3}}{3 x ^{2} + 1}

= \int \frac{x ^{3}}{3 x ^{2} + 1} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{x ^{3}}{x ^{2} + 1} d x

Apply u-substitution

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

= \frac{1}{3} \cdot \int u^{2} - 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

= \frac{1}{3} (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

= \frac{1}{3} (\frac{u ^{3}}{3} - u)

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x ^{3}}{x ^{8} + 4} d x

\int e^{- 6 x + 5} d x

\int 2 y^{4} d y

\int \frac{1}{( 3 x + 5 ) ^{2}} d x

\int \frac{x + 4}{x ^{2}} d x

What is the integral of (x^5)/(x^2sqrt(9x^2+9)) ?
The integral of (x^5)/(x^2sqrt(9x^2+9)) is 1/3 (1/3 (x^2+1)^{3/2}-sqrt(x^2+1))+C