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

The integral of xsqrt(1+4x^2) is 1/12 (1+4x^2)^{3/2}+C

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integral of xsqrt(1+4x^2)

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

\frac{1}{12} (1 + 4 x^{2})^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{8} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 1 + 4 x^{2}

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

Simplify

\frac{1}{8} \cdot \frac{2}{3} (1 + 4 x^{2})^{\frac{3}{2}} : \frac{1}{12} (1 + 4 x^{2})^{\frac{3}{2}}

= \frac{1}{12} (1 + 4 x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (1 + 4 x^{2})^{\frac{3}{2}} + C

x y^{^{'}} - 2 y = 2

\int_{4}^{\infty} e^{- x} d x

f (x) = 2 x e^{x} csc (x)

\frac{d}{d x} (sec^{2} (\frac{x}{2}))

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

What is the integral of xsqrt(1+4x^2) ?
The integral of xsqrt(1+4x^2) is 1/12 (1+4x^2)^{3/2}+C