What is the integral of t/(sqrt(t^4-4)) ?

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

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integral of t/(sqrt(t^4-4))

\int \frac{t}{t ^{4} - 4} d t

\frac{1}{2} ln (\frac{t ^{4} - 4 + t ^{2}}{2}) + C

Solution steps

\int \frac{t}{t ^{4} - 4} d t

Apply u-substitution

: \int \frac{1}{2 u ^{2} - 4} d u

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \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 (arcsec (\frac{1}{2} t^{2})) + sec (arcsec (\frac{1}{2} t^{2}))

Simplify

\frac{1}{2} ln tan (arcsec (\frac{1}{2} t^{2})) + sec (arcsec (\frac{1}{2} t^{2})) : \frac{1}{2} ln (\frac{t ^{4} - 4 + t ^{2}}{2})

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

Add a constant to the solution

= \frac{1}{2} ln (\frac{t ^{4} - 4 + t ^{2}}{2}) + C

\int [3 e^{x}]^{4} d x

\frac{\partial}{\partial x} (x^{2} e^{5 x y})

x \to 0 - lim (csc (x) - cot (x))

\int \frac{6}{z ^{4}} d z

x \to 4 lim (\frac{x}{x})

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