What is the integral of 2xsqrt(4x^2+1) ?

The integral of 2xsqrt(4x^2+1) is 1/6 (4x^2+1)^{3/2}+C

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integral of 2xsqrt(4x^2+1)

\int 2 x 4 x^{2} + 1 d x

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

Solution steps

\int 2 x 4 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 4 x^{2} + 1 d x

Apply u-substitution

= 2 \cdot \int \frac{u}{8} d u

Take the constant out:

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

= 2 \cdot \frac{1}{8} \cdot \int u d u

Apply the Power Rule

= 2 \cdot \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 x^{2} + 1

= 2 \cdot \frac{1}{8} \cdot \frac{2}{3} (4 x^{2} + 1)^{\frac{3}{2}}

Simplify

2 \cdot \frac{1}{8} \cdot \frac{2}{3} (4 x^{2} + 1)^{\frac{3}{2}} : \frac{1}{6} (4 x^{2} + 1)^{\frac{3}{2}}

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

Add a constant to the solution

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

n = 0 \sum \infty \frac{2}{n ^{2} + 4 n + 3}

\int x - x^{2} d x

\frac{d}{d x} (\frac{- 4 x ^{6} + 4 x ^{5} - 3 x ^{3}}{3 x ^{2}})

\int e^{2 s i n^{2} (x)} sin (2 x) d x

d er i v a t i v e \frac{50}{x}

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