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

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

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integral of 7/(1-x^2)

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

7 (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}) + C

Solution steps

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{1 - x ^{2}} d x = \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}

= 7 (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})

Add a constant to the solution

= 7 (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}) + C

(2 t - 1) (5 t - 5)^{- 1}

x \to 2 lim (\frac{1}{( x - 2 ) ^{3}})

y^{^{''}} - y^{^{'}} - 2 y = - 2 t + 8 t^{2}

\frac{4}{( s - 1 ) ^{3}}

y^{^{'}} = a - b y^{2}

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