What is the integral of sqrt(x^2+1-1/2+1/(16x^2)) ?

The integral of sqrt(x^2+1-1/2+1/(16x^2)) is 1/4 (2x^2+ln|x|)+C

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integral of sqrt(x^2+1-1/2+1/(16x^2))

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

\frac{1}{4} (2 x^{2} + ln ∣ x ∣) + C

Solution steps

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

Simplify

x^{2} + 1 - \frac{1}{2} + \frac{1}{16 x ^{2}} : \frac{4 x ^{2} + 1}{4 x}

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

Take the constant out:

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

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

Expand

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{1}{4} (2 x^{2} + ln ∣ x ∣)

Add a constant to the solution

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

\int \frac{cos ( 2 x )}{cos ( x ) - sin ( x )} d x

\frac{d}{d x} (5 cos^{- 4} (x))

\int \frac{e ^{2 x}}{e ^{2 x} - 1} d x

\frac{d}{d x} (sin (x + π))

x^{2} e^{x}

What is the integral of sqrt(x^2+1-1/2+1/(16x^2)) ?
The integral of sqrt(x^2+1-1/2+1/(16x^2)) is 1/4 (2x^2+ln|x|)+C