What is the integral of x^3sqrt(x^4+10) ?

The integral of x^3sqrt(x^4+10) is 1/6 (x^4+10)^{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^4+10)

\int x^{3} x^{4} + 10 d x

\frac{1}{6} (x^{4} + 10)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{4} + 10 d x

Apply u-substitution

: \int \frac{u}{4} d u

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

Take the constant out:

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

= \frac{1}{4} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x^{4} + 10

= \frac{1}{4} \cdot \frac{2}{3} (x^{4} + 10)^{\frac{3}{2}}

Simplify

\frac{1}{4} \cdot \frac{2}{3} (x^{4} + 10)^{\frac{3}{2}} : \frac{1}{6} (x^{4} + 10)^{\frac{3}{2}}

= \frac{1}{6} (x^{4} + 10)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (x^{4} + 10)^{\frac{3}{2}} + C

4 y^{^{''}} + 4 y^{^{'}} + 9 y = 0

\int_{2}^{3} \frac{1}{x ^{2} - 1} d x

\int (y - 2)^{2} d y

\frac{d}{d x} (\frac{( x ^{2} - 5 ) ^{3}}{x ^{2} + 4})

\int + 4 e^{- 7 x} d x

What is the integral of x^3sqrt(x^4+10) ?
The integral of x^3sqrt(x^4+10) is 1/6 (x^4+10)^{3/2}+C