What is the integral of 1/(xsqrt(49x^4-4)) ?

The integral of 1/(xsqrt(49x^4-4)) is 1/4 arctan(1/2 sqrt(49x^4-4))+C

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integral of 1/(xsqrt(49x^4-4))

\int \frac{1}{x 49 x ^{4} - 4} d x

\frac{1}{4} arctan (\frac{1}{2} 49 x^{4} - 4) + C

Solution steps

\int \frac{1}{x 49 x ^{4} - 4} d x

Apply u-substitution

: \int \frac{1}{4 u 49 u - 4} d u

= \int \frac{1}{4 u 49 u - 4} 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 49 u - 4} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 4} d v

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

= \frac{1}{4} \cdot 2 \cdot \frac{1}{2} arctan (\frac{49 x ^{4} - 4}{2})

Simplify

\frac{1}{4} \cdot 2 \cdot \frac{1}{2} arctan (\frac{49 x ^{4} - 4}{2}) : \frac{1}{4} arctan (\frac{1}{2} 49 x^{4} - 4)

= \frac{1}{4} arctan (\frac{1}{2} 49 x^{4} - 4)

Add a constant to the solution

= \frac{1}{4} arctan (\frac{1}{2} 49 x^{4} - 4) + C

y^{^{''}} - y^{^{'}} - 2 y = t e^{2 t}

- \frac{1}{9} (x^{- 9} - x^{18})

\int ln (9 x + 1) d x

\frac{d}{d x} (3 x + 8)

t (\frac{d y}{d t}) = 2 t e^{t} - y + 6 t^{2}

What is the integral of 1/(xsqrt(49x^4-4)) ?
The integral of 1/(xsqrt(49x^4-4)) is 1/4 arctan(1/2 sqrt(49x^4-4))+C