What is the integral of (6x)/(1-7x^2) ?

The integral of (6x)/(1-7x^2) is -3/7 ln|1-7x^2|+C

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integral of (6x)/(1-7x^2)

\int \frac{6 x}{1 - 7 x ^{2}} d x

- \frac{3}{7} ln 1 - 7 x^{2} + C

Solution steps

\int \frac{6 x}{1 - 7 x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{1 - 7 x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{14 u} d u

= 6 \cdot \int - \frac{1}{14 u} d u

Take the constant out:

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

= 6 (- \frac{1}{14} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 (- \frac{1}{14} ln ∣ u ∣)

Substitute back

u = 1 - 7 x^{2}

= 6 (- \frac{1}{14} ln 1 - 7 x^{2})

Simplify

6 (- \frac{1}{14} ln 1 - 7 x^{2}) : - \frac{3}{7} ln 1 - 7 x^{2}

= - \frac{3}{7} ln 1 - 7 x^{2}

Add a constant to the solution

= - \frac{3}{7} ln 1 - 7 x^{2} + C

2 x y^{^{'}} + y = 10 x

\frac{d}{d x} (\frac{2 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}})

\int (x^{3} + 1)^{3} d x

f (x) = \frac{8}{7 x ^{2}}

What is the integral of (6x)/(1-7x^2) ?
The integral of (6x)/(1-7x^2) is -3/7 ln|1-7x^2|+C