What is the integral of x/(sqrt(x^4+36)) ?

The integral of x/(sqrt(x^4+36)) is 1/2 ln((|x^2+sqrt(36+x^4)|)/6)+C

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integral of x/(sqrt(x^4+36))

\int \frac{x}{x ^{4} + 36} d x

\frac{1}{2} ln (\frac{x ^{2} + 36 + x ^{4}}{6}) + C

Solution steps

\int \frac{x}{x ^{4} + 36} d x

Apply u-substitution

= \int \frac{1}{2 u ^{2} + 36} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 36} d u

Apply Trigonometric Substitution

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

Use the common integral:

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

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{2} ln tan (arctan (\frac{1}{6} x^{2})) + sec (arctan (\frac{1}{6} x^{2}))

Simplify

\frac{1}{2} ln tan (arctan (\frac{1}{6} x^{2})) + sec (arctan (\frac{1}{6} x^{2})) : \frac{1}{2} ln (\frac{x ^{2} + 36 + x ^{4}}{6})

= \frac{1}{2} ln (\frac{x ^{2} + 36 + x ^{4}}{6})

Add a constant to the solution

= \frac{1}{2} ln (\frac{x ^{2} + 36 + x ^{4}}{6}) + C

\int \frac{25}{x ^{3} - 27} d x

n = 1 \sum \infty \frac{( n ^{3} )}{3 ^{n^{2}}}

y^{^{''}} - 4 y^{^{'}} + 7 y = t e^{t}

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

x \to 0 + lim ((1 + x)^{\frac{5}{x}})

What is the integral of x/(sqrt(x^4+36)) ?
The integral of x/(sqrt(x^4+36)) is 1/2 ln((|x^2+sqrt(36+x^4)|)/6)+C