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

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

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

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

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

Solution steps

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

Simplify

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

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

x^{4} = x^{2},

assuming

x \geq 0

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

x x^{2} = x^{3}

= \int \frac{x ^{4} + 1}{x ^{3}} d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 32 x e^{x}

\int x^{5} cos (x) d x

f (x) = \frac{6 x}{x ^{2} + 1}

\frac{d}{d x} (3 x (18 x^{4} + \frac{13}{x + 1}))

\frac{d}{d x} (ln (sin (x^{3})))

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