What is the integral of sqrt(1+(2x-1/(8x))^2) ?

The integral of sqrt(1+(2x-1/(8x))^2) is x^2+1/8 ln|x|+C

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integral of sqrt(1+(2x-1/(8x))^2)

\int 1 + (2 x - \frac{1}{8 x})^{2} d x

x^{2} + \frac{1}{8} ln ∣ x ∣ + C

Solution steps

\int 1 + (2 x - \frac{1}{8 x})^{2} d x

Simplify

1 + (2 x - \frac{1}{8 x})^{2} : \frac{16 x ^{2} + 1}{8 x}

= \int \frac{16 x ^{2} + 1}{8 x} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{16 x ^{2} + 1}{x} d x

Expand

\frac{16 x ^{2} + 1}{x} : 16 x + \frac{1}{x}

= \frac{1}{8} \cdot \int 16 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}{8} (\int 16 x d x + \int \frac{1}{x} d x)

\int 16 x d x = 8 x^{2}

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{1}{8} (8 x^{2} + ln ∣ x ∣)

Simplify

\frac{1}{8} (8 x^{2} + ln ∣ x ∣) : x^{2} + \frac{1}{8} ln ∣ x ∣

= x^{2} + \frac{1}{8} ln ∣ x ∣

Add a constant to the solution

= x^{2} + \frac{1}{8} ln ∣ x ∣ + C

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

(e^{12 x})^{^{'}}

\int sinh^{2} (x) cosh (x) d x

x \to 0 lim (x cos (\frac{1}{x ^{2}}))

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

What is the integral of sqrt(1+(2x-1/(8x))^2) ?
The integral of sqrt(1+(2x-1/(8x))^2) is x^2+1/8 ln|x|+C