What is the integral of (7x^3)/(sqrt(3+x^2)) ?

The integral of (7x^3)/(sqrt(3+x^2)) is 7(1/3 (3+x^2)^{3/2}-3sqrt(3+x^2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (7x^3)/(sqrt(3+x^2))

\int \frac{7 x ^{3}}{3 + x ^{2}} d x

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

Solution steps

\int \frac{7 x ^{3}}{3 + x ^{2}} d x

Take the constant out:

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

= 7 \cdot \int \frac{x ^{3}}{3 + x ^{2}} d x

Apply u-substitution

: \int u^{2} - 3 d u

= 7 \cdot \int u^{2} - 3 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

= 7 (\int u^{2} d u - \int 3 d u)

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

\int 3 d u = 3 u

= 7 (\frac{u ^{3}}{3} - 3 u)

Substitute back

u = 3 + x^{2}

= 7 (\frac{( 3 + x ^{2} ) ^{3}}{3} - 3 3 + x^{2})

Simplify

7 (\frac{( 3 + x ^{2} ) ^{3}}{3} - 3 3 + x^{2}) : 7 (\frac{1}{3} (3 + x^{2})^{\frac{3}{2}} - 3 3 + x^{2})

= 7 (\frac{1}{3} (3 + x^{2})^{\frac{3}{2}} - 3 3 + x^{2})

Add a constant to the solution

= 7 (\frac{1}{3} (3 + x^{2})^{\frac{3}{2}} - 3 3 + x^{2}) + C

\frac{8}{9 s ^{2} + 9 s + 4}

\frac{x ^{2} + 32 x}{8}

\frac{8}{s - 1} - \frac{3}{s}

(2 x^{2}) + (x - 8)

x \to 5 lim (\frac{x + 5}{x - 5})

What is the integral of (7x^3)/(sqrt(3+x^2)) ?
The integral of (7x^3)/(sqrt(3+x^2)) is 7(1/3 (3+x^2)^{3/2}-3sqrt(3+x^2))+C