What is the integral of 2/(tsqrt(4t^4-1)) ?

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

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integral of 2/(tsqrt(4t^4-1))

\int \frac{2}{t 4 t ^{4} - 1} d t

arctan (4 t^{4} - 1) + C

Solution steps

\int \frac{2}{t 4 t ^{4} - 1} d t

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{4} \cdot 2 arctan (v)

Substitute back

= 2 \cdot \frac{1}{4} \cdot 2 arctan (4 t^{4} - 1)

Simplify

2 \cdot \frac{1}{4} \cdot 2 arctan (4 t^{4} - 1) : arctan (4 t^{4} - 1)

= arctan (4 t^{4} - 1)

Add a constant to the solution

= arctan (4 t^{4} - 1) + C

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