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

The integral of x/(sqrt(1+4x^2)) is 1/4 sqrt(1+4x^2)+C

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integral of x/(sqrt(1+4x^2))

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

\frac{1}{4} 1 + 4 x^{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{8 u} d u

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 1 + 4 x^{2}

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

Simplify

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

= \frac{1}{4} 1 + 4 x^{2}

Add a constant to the solution

= \frac{1}{4} 1 + 4 x^{2} + C

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

\int_{0}^{1} \frac{65}{x ^{5}} d x

y^{^{'}} + 4 x y = 0

\int 4 (2 x + 5)^{3} d x

1 - x^{2}, x = 0

What is the integral of x/(sqrt(1+4x^2)) ?
The integral of x/(sqrt(1+4x^2)) is 1/4 sqrt(1+4x^2)+C