What is the integral of (3x)/(sqrt(x^4+6x^2+5)) ?

The integral of (3x)/(sqrt(x^4+6x^2+5)) is 3/2 ln| 1/(sqrt(11))x^2+sqrt(1/11 (11+x^4))|+C

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integral of (3x)/(sqrt(x^4+6x^2+5))

\int \frac{3 x}{x ^{4} + 6 x ^{2} + 5} d x

\frac{3}{2} ln \frac{1}{11} x^{2} + \frac{1}{11} (11 + x^{4}) + C

Solution steps

\int \frac{3 x}{x ^{4} + 6 x ^{2} + 5} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{4} + 6 x ^{2} + 5} d x

Apply u-substitution

= 3 \cdot \int \frac{1}{4 u 6 u + u + 5} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int \frac{1}{u 6 u + u + 5} d u

Apply exponent rule:

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

\frac{1}{u 6 u + u + 5} = u^{- \frac{1}{2}} \frac{1}{6 + u + 5}

= 3 \cdot \frac{1}{4} \cdot \int u^{- \frac{1}{2}} \frac{1}{6 + u + 5} d u

Simplify

u^{- \frac{1}{2}} \frac{1}{6 + u + 5} : \frac{1}{u u + 11}

= 3 \cdot \frac{1}{4} \cdot \int \frac{1}{u u + 11} d u

Apply u-substitution

= 3 \cdot \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 11} d v

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 11} d v

Apply Trigonometric Substitution

= 3 \cdot \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) ∣

= 3 \cdot \frac{1}{4} \cdot 2 ln ∣ tan (w) + sec (w) ∣

Substitute back

= 3 \cdot \frac{1}{4} \cdot 2 ln tan (arctan (\frac{1}{11} x^{4})) + sec (arctan (\frac{1}{11} x^{4}))

Simplify

3 \cdot \frac{1}{4} \cdot 2 ln tan (arctan (\frac{1}{11} x^{4})) + sec (arctan (\frac{1}{11} x^{4})) : \frac{3}{2} ln \frac{1}{11} x^{2} + \frac{1}{11} (11 + x^{4})

= \frac{3}{2} ln \frac{1}{11} x^{2} + \frac{1}{11} (11 + x^{4})

Add a constant to the solution

= \frac{3}{2} ln \frac{1}{11} x^{2} + \frac{1}{11} (11 + x^{4}) + C

What is the integral of (3x)/(sqrt(x^4+6x^2+5)) ?
The integral of (3x)/(sqrt(x^4+6x^2+5)) is 3/2 ln| 1/(sqrt(11))x^2+sqrt(1/11 (11+x^4))|+C