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

The integral of xsqrt(2x^2+1) is 1/6 (2x^2+1)^{3/2}+C

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integral of xsqrt(2x^2+1)

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (x \cdot arctan (x))

\int (\frac{1}{2} (1 - cos (2 x)))^{2} d x

x \to - 0.1 lim (x^{3} + 6 x^{2} + x)

\frac{d}{d x} (e^{2 x - 1})

\int \frac{3 x ^{2} - 8 x + 1}{x ^{3} - 4 x ^{2} + x + 6} d x

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