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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u 1 + u} d u

= \int \frac{1}{4 u 1 + u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u 1 + u} d u

Apply u-substitution

: \int \frac{2}{v ^{2} - 1} d v

= \frac{1}{4} \cdot \int \frac{2}{v ^{2} - 1} d v

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{4} \cdot 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{d y}{d x} = \frac{7 x + 2 y}{7 x + 2 y + 2}

f (x) = 2 - 7 x, (- 2, 4)

\frac{d}{d x} (∣ cos (x) ∣)

(y^{3} - 1) e^{x} d x + 3 y^{2} (e^{x} + 1) d y = 0, y (0) = - 2

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