What is the integral of x/(sqrt(x^4-16x^2+39)) ?

The integral of x/(sqrt(x^4-16x^2+39)) is 1/2 ln| 1/(sqrt(23))x^2+sqrt(1/23 (23+x^4))|+C

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integral of x/(sqrt(x^4-16x^2+39))

\int \frac{x}{x ^{4} - 16 x ^{2} + 39} d x

\frac{1}{2} ln \frac{1}{23} x^{2} + \frac{1}{23} (23 + x^{4}) + C

Solution steps

\int \frac{x}{x ^{4} - 16 x ^{2} + 39} d x

Apply u-substitution

: \int \frac{1}{4 u u + 39 - 16 u} d u

= \int \frac{1}{4 u u + 39 - 16 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 u + 39 - 16 u} d u

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{u u + 39 - 16 u} = u^{- \frac{1}{2}} \frac{1}{u + 39 - 16}

= \frac{1}{4} \cdot \int u^{- \frac{1}{2}} \frac{1}{u + 39 - 16} d u

Simplify

u^{- \frac{1}{2}} \frac{1}{u + 39 - 16} : \frac{1}{u u + 23}

= \frac{1}{4} \cdot \int \frac{1}{u u + 23} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 23} d v

= \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 23} 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} + 23} d v

Apply Trigonometric Substitution

: \int sec (w) d w

= \frac{1}{4} \cdot 2 \cdot \int sec (w) d w

Use the common integral:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= \frac{1}{4} \cdot 2 ln ∣ tan (w) + sec (w) ∣

Substitute back

= \frac{1}{4} \cdot 2 ln tan (arctan (\frac{1}{23} x^{4})) + sec (arctan (\frac{1}{23} x^{4}))

Simplify

\frac{1}{4} \cdot 2 ln tan (arctan (\frac{1}{23} x^{4})) + sec (arctan (\frac{1}{23} x^{4})) : \frac{1}{2} ln \frac{1}{23} x^{2} + \frac{1}{23} (23 + x^{4})

= \frac{1}{2} ln \frac{1}{23} x^{2} + \frac{1}{23} (23 + x^{4})

Add a constant to the solution

= \frac{1}{2} ln \frac{1}{23} x^{2} + \frac{1}{23} (23 + x^{4}) + C

y = x + x^{2} + x^{3}, x = 1

\int_{0}^{e} a^{x} d x

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\int - 3 cos (2 x) d x

\int \frac{2 ^{x}}{x} d x

What is the integral of x/(sqrt(x^4-16x^2+39)) ?
The integral of x/(sqrt(x^4-16x^2+39)) is 1/2 ln| 1/(sqrt(23))x^2+sqrt(1/23 (23+x^4))|+C