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

The integral of sqrt(1+((x-1)^2)/(4x)) is 1/3 sqrt(x)(x+3)+C

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

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

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

Solution steps

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

Simplify

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

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{( x + 1 ) ^{2}}{x} d x

(x + 1)^{2} = ((x + 1)),

assuming

(x + 1) \geq 0

= \frac{1}{2} \cdot \int \frac{x + 1}{x} d x

Expand

\frac{x + 1}{x} : x + \frac{1}{x}

= \frac{1}{2} \cdot \int x + \frac{1}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int x d x + \int \frac{1}{x} d x)

\int x d x = \frac{2}{3} x^{\frac{3}{2}}

\int \frac{1}{x} d x = 2 x

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

Simplify

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

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

Add a constant to the solution

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

x \to - \infty lim (2 e^{x})

\frac{d}{d x} (A x + B)

4 d y + y d x = x^{2} d y

y^{^{'}} = 2 x - 3 y + 1

e^{2 y} y^{^{'}} = 3

What is the integral of sqrt(1+((x-1)^2)/(4x)) ?
The integral of sqrt(1+((x-1)^2)/(4x)) is 1/3 sqrt(x)(x+3)+C