What is the integral of 2x^3sqrt(x^2+1) ?

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{( u - 1 ) u}{2} d u

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

Take the constant out:

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

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

Expand

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

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

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= 2 \cdot \frac{1}{2} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{1 - e ^{2 x}} d x

\frac{d}{d x} (2 lo g_{6} (x))

y^{^{''}} + 3 y^{^{'}} + 2.25 y = 0

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

\frac{d y}{d t} = 7 y^{2} t^{3}, y (2) = 3

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