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

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

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integral of (2sqrt(1+x^4))/(x^3)

\int \frac{2 1 + x ^{4}}{x ^{3}} d x

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

Solution steps

\int \frac{2 1 + x ^{4}}{x ^{3}} d x

Take the constant out:

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

= 2 \cdot \int \frac{1 + x ^{4}}{x ^{3}} d x

Apply Integration By Parts

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

x^{^{''}} - x^{^{'}} = 0

y - 2 y^{^{''}} + y^{^{''''}} = 0

x \to - 2 lim (lo g_{2} (8))

(x, y) \to (0, 0) lim (x - y)

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