What is the integral of (9xsqrt(1+2x^2)) ?

The integral of (9xsqrt(1+2x^2)) is 3/2 (2x^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 (9xsqrt(1+2x^2))

\int (9 x 1 + 2 x^{2}) d x

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

Solution steps

\int 9 x 1 + 2 x^{2} d x

Take the constant out:

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

= 9 \cdot \int x 1 + 2 x^{2} d x

Apply u-substitution

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

= 9 \cdot \int \frac{u}{4} d u

Take the constant out:

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

= 9 \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= 9 \cdot \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 2 x^{2}

= 9 \cdot \frac{1}{4} \cdot \frac{2}{3} (1 + 2 x^{2})^{\frac{3}{2}}

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim (\frac{x ^{2} - x}{x})

\frac{d}{d x} (\frac{x}{x ^{2} + y ^{2}})

\int_{0}^{1} x - x^{4} d x

y^{^{''}} + 4 y^{^{'}} - 2 y = 0

\frac{\partial}{\partial x} (\frac{x}{x ^{3} - y ^{6}})

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