What is the integral of 3/(x^{-1)+1} ?

The integral of 3/(x^{-1)+1} is 3(-ln|1+x|+x)+C

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integral of 3/(x^{-1)+1}

\int \frac{3}{x ^{- 1} + 1} d x

3 (- ln ∣ 1 + x ∣ + x) + C

Solution steps

\int \frac{3}{x ^{- 1} + 1} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{- 1} + 1} d x

\frac{1}{x ^{- 1} + 1} = - \frac{1}{1 + x} + 1

= 3 \cdot \int - \frac{1}{1 + x} + 1 d x

Apply the Sum Rule:

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

= 3 (- \int \frac{1}{1 + x} d x + \int 1 d x)

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

\int 1 d x = x

= 3 (- ln ∣ 1 + x ∣ + x)

Add a constant to the solution

= 3 (- ln ∣ 1 + x ∣ + x) + C

6 + 5 x

∣ x ∣, x^{2} - 2

1 - 9 x^{2} arccos (3 x)

n = 2 \sum \infty n

f (t) = e^{3 t s i n (2 t)}

What is the integral of 3/(x^{-1)+1} ?
The integral of 3/(x^{-1)+1} is 3(-ln|1+x|+x)+C