What is the integral of 1/((7x-1)sqrt(49x^2-14x)) ?

The integral of 1/((7x-1)sqrt(49x^2-14x)) is 1/7 arctan(sqrt(49x^2-14x))+C

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integral of 1/((7x-1)sqrt(49x^2-14x))

\int \frac{1}{( 7 x - 1 ) 49 x ^{2} - 14 x} d x

\frac{1}{7} arctan (49 x^{2} - 14 x) + C

Solution steps

\int \frac{1}{( 7 x - 1 ) 49 x ^{2} - 14 x} d x

Apply u-substitution

: \int \frac{1}{7 u u ^{2} - 1} d u

= \int \frac{1}{7 u u ^{2} - 1} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{1}{u u ^{2} - 1} d u

Apply u-substitution

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

= \frac{1}{7} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{7} arctan (v)

Substitute back

= \frac{1}{7} arctan ((7 x - 1)^{2} - 1)

Simplify

\frac{1}{7} arctan ((7 x - 1)^{2} - 1) : \frac{1}{7} arctan (49 x^{2} - 14 x)

= \frac{1}{7} arctan (49 x^{2} - 14 x)

Add a constant to the solution

= \frac{1}{7} arctan (49 x^{2} - 14 x) + C

\int_{1}^{2} e^{\frac{1}{x}} d x

\int (\frac{2 y}{( x ^{2} + y ^{2} ) ^{2}}) d y

f (x) = 4 x^{- \frac{3}{8}}

\frac{d}{d x} (\frac{x ^{7}}{7} - x^{3} - \frac{3}{x} + \frac{1}{5 x ^{5}})

x \to 0 lim (\frac{sin ( 3 x ^{2} )}{x})

What is the integral of 1/((7x-1)sqrt(49x^2-14x)) ?
The integral of 1/((7x-1)sqrt(49x^2-14x)) is 1/7 arctan(sqrt(49x^2-14x))+C