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

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

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

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

Solution steps

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

Apply u-substitution

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

= 2 \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

= 2 \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = 2 x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int 31.863 x^{- 0.61} d x

\int 12 (y^{4} + 4 y^{2} + 7)^{2} (y^{3} + 2 y) d y

\int x 4 x^{2} + 11 d x

\int \frac{5 x}{x ^{4} - 1} d x

\int (x^{8.3} - 9 x^{6} + 3 x^{- 4} + x^{- 3}) d x

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