# Thread: How to integrate (x+1)/(2x+3)

1. ## How to integrate (x+1)/(2x+3)

Hello,

What is the method of integrating I = (x + 1)/(2x + 3)?

I get a wrong answer: 0.5x - 0.5ln(2x + 3)
The correct answer is 0.5x - 0.25ln(2x + 3)

My method:

Let (x+1) = A{2x + 3} + B{differential coefficient of 2x + 3}

(x+1) = A{2x + 3} + B{2}

equating coefficient of x: 1 = 2A :. A = 0.5
constants: 1 = 3A + B :. B = -0.5

:. I = 0.5x - 0.5ln(2x + 3)

Thank you

2. Originally Posted by algorithm
Hello,

What is the method of integrating I = (x + 1)/(2x + 3)?

I get a wrong answer: 0.5x - 0.5ln(2x + 3)
The correct answer is 0.5x - 0.25ln(2x + 3)

My method:

Let (x+1) = A{2x + 3} + B{differential coefficient of 2x + 3}

(x+1) = A{2x + 3} + B{2}

equating coefficient of x: 1 = 2A :. A = 0.5
constants: 1 = 3A + B :. B = -0.5

:. I = 0.5x - 0.5ln(2x + 3)

Thank you
several approaches here.

you can do long division to find $x + 1 \div 2x + 3$ to get $\frac 12 \left( 1 - \frac 1{2x + 3} \right)$ ...which should be easy to integrate

or just use algebra:

note that $\frac {x + 1}{2x + 3} = \frac 12 \left( \frac {2x + 2}{2x + 3} \right) = \frac 12 \left( \frac {2x + 3 - 1}{2x + 3} \right) = \frac 12 \left(1 - \frac {1}{2x + 3} \right)$

3. It's also possible to substitute:
$\int\frac{x+1}{2x+3}\,\text{d}x=\left|\begin{array }{rcl}2x+3&=&\varphi\\x&=&\dfrac{\varphi-3}{2}\\\text{d}x&=&\frac{1}{2}\,\text{d}\varphi\en d{array}\right|=\frac{1}{4}\int\frac{\varphi-1}{\varphi}\,\text{d}\varphi=\frac{1}{4}\int\left( 1-\frac{1}{\varphi}\right)\,\text{d}\varphi$
So the solution is function $F(x)=\frac{2x+3}{4}-\frac{1}{4}\ln|2x+3|+C$

4. Originally Posted by algorithm
My method:

Let (x+1) = A{2x + 3} + B{differential coefficient of 2x + 3}

(x+1) = A{2x + 3} + B{2}

equating coefficient of x: 1 = 2A :. A = 0.5
constants: 1 = 3A + B :. B = -0.5

:. I = 0.5x - 0.5ln(2x + 3)
The problem is that $\ln(2x+3)$ is not an antiderivative of $\frac{1}{2x+3}$. Do you see why this is so ?

5. Thanks for the replies.