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

The integral of ((2x-1)^2)/(2x) is x^2-2x+1/2 ln|x|+C

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integral of ((2x-1)^2)/(2x)

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

x^{2} - 2 x + \frac{1}{2} ln ∣ x ∣ + C

Solution steps

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

Expand

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

= \frac{1}{2} \cdot \int 4 x - 4 + \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 4 x d x - \int 4 d x + \int \frac{1}{x} d x)

\int 4 x d x = 2 x^{2}

\int 4 d x = 4 x

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{1}{2} (2 x^{2} - 4 x + ln ∣ x ∣)

Simplify

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

= x^{2} - 2 x + \frac{1}{2} ln ∣ x ∣

Add a constant to the solution

= x^{2} - 2 x + \frac{1}{2} ln ∣ x ∣ + C

\int_{0}^{\frac{1}{2}} 1 d x

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

\frac{d y}{d x} + y cos (x) = 2 cos (x), y (0) = 4

f (x) = sin (x) ln (2 x)

sec (θ)

What is the integral of ((2x-1)^2)/(2x) ?
The integral of ((2x-1)^2)/(2x) is x^2-2x+1/2 ln|x|+C