# Find the antiderivative of the function f(x)=(x+1)/(x^2+2x) .

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### 2 Answers

We have to find the anti derivative of f(x)=(x+1)/(x^2+2x)

f(x)=(x+1)/(x^2+2x)

let x^2 + 2x = t

=> dt/dx = 2x + 2 = 2(x + 1)

Int [(x+1)/(x^2+2x) dx]

=> Int [ (1/2t) dt]

=> (1/2) ln t + C

replace t with x^2 + 2x

=> (1/2) ln (x^2 + 2x) + C

**The required result is (1/2) ln (x^2 + 2x) + C**

We'll determine the indefinite integral of the given function:

Int f(x)dx = Int (x+1)dx/(x^2+2x)

We notice that if we'll differentiate the denominator of the function, we'll get the numerator multiplied by 2.

We'll substitute the denominator by t.

x^2+2x = t

We'll differentiate both sides:

(2x + 2)dx = dt

We'll divide by 2:

(x + 1)dx = dt/2

We'll re-write the integral in t:

Int f(x)dx = Int dt/2t = (1/2)*ln |t| + C

Int f(x)dx = (1/2)*ln |x^2+2x| + C

**Int f(x)dx = ln sqrt (x^2+2x) + C**