What is the integral of (3+u)/(1-u^2) ?

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

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integral of (3+u)/(1-u^2)

\int \frac{3 + u}{1 - u ^{2}} d u

3 (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}) - \frac{1}{2} ln 1 - u^{2} + C

Solution steps

\int \frac{3 + u}{1 - u ^{2}} d u

Expand

\frac{3 + u}{1 - u ^{2}} : \frac{3}{1 - u ^{2}} + \frac{u}{1 - u ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3}{1 - u ^{2}} d u + \int \frac{u}{1 - u ^{2}} d u

\int \frac{3}{1 - u ^{2}} d u = 3 (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

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

= 3 (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}) - \frac{1}{2} ln 1 - u^{2}

Add a constant to the solution

= 3 (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}) - \frac{1}{2} ln 1 - u^{2} + C

x \to - 5 lim (x^{2} + 5)

\frac{d y}{d x} = e^{y - 2 x}

\int (7 sec (x) tan (x) + 6 sec^{2} (x)) d x

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