What is the integral of x^3sqrt(x^2+17) ?

The integral of x^3sqrt(x^2+17) is 1/5 (x^2+17)^{5/2}-17/3 (x^2+17)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^3sqrt(x^2+17)

\int x^{3} x^{2} + 17 d x

\frac{1}{5} (x^{2} + 17)^{\frac{5}{2}} - \frac{17}{3} (x^{2} + 17)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 17 d x

Apply u-substitution

: \int (u^{2} - 17) u^{2} d u

= \int (u^{2} - 17) u^{2} d u

Expand

(u^{2} - 17) u^{2} : u^{4} - 17 u^{2}

= \int u^{4} - 17 u^{2} 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

= \int u^{4} d u - \int 17 u^{2} d u

\int u^{4} d u = \frac{u ^{5}}{5}

\int 17 u^{2} d u = \frac{17 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{17 u ^{3}}{3}

Substitute back

u = x^{2} + 17

= \frac{( x ^{2} + 17 ) ^{5}}{5} - \frac{17 ( x ^{2} + 17 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 17 ) ^{5}}{5} - \frac{17 ( x ^{2} + 17 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 17)^{\frac{5}{2}} - \frac{17}{3} (x^{2} + 17)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 17)^{\frac{5}{2}} - \frac{17}{3} (x^{2} + 17)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 17)^{\frac{5}{2}} - \frac{17}{3} (x^{2} + 17)^{\frac{3}{2}} + C

\frac{d v}{d t} = k \cdot (275 - v), v (0) = 0

cos^{2} (x) + sin^{2} (x)

y = ln (x^{2} - 9 x + 1), (9, 0)

\int 5 sin (t) d t

\frac{0.625}{s + 4}

What is the integral of x^3sqrt(x^2+17) ?
The integral of x^3sqrt(x^2+17) is 1/5 (x^2+17)^{5/2}-17/3 (x^2+17)^{3/2}+C