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

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

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

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

ln x^{4} + 1 + x^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 1} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = ln u^{2} + 1 + u

= 2 \cdot \frac{1}{2} ln u^{2} + 1 + u

Substitute back

u = x^{2}

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

Simplify

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

= ln x^{4} + 1 + x^{2}

Add a constant to the solution

= ln x^{4} + 1 + x^{2} + C

\int 5 x^{2} a^{2} - x^{2} d x

x \to \frac{π}{6} lim (cot (2 x))

x \frac{d y}{d x} - y = 2020 x^{2}

\int_{0}^{x^{2}} sin (t) d t

\frac{5}{x}, \frac{5}{x ^{2}} 1, 7

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