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## Integral Calculus

### Course: Integral Calculus>Unit 1

Lesson 18: Integrating using linear partial fractions

# Integration with partial fractions

Finding the integral of a rational function using linear partial fraction decomposition.

## Want to join the conversation?

• I think that at the end of the vedio, the answer of that question should be -7In|2x-3|+4In|x-1| instead of -7/2|2x-3|+4In|x-1| because the numerator's 2 can be cancel out. Can anyone help me with that?
• Sure, it's because of the chain rule. Remember that the derivative of 2x-3 is 2, thus to take the integral of 1/(2x-3), we must include a factor of 1/2 outside the integral so that the inside becomes 2/(2x-3), which has an antiderivative of ln(2x+3). Again, this is because the derivative of ln(2x+3) is 1/(2x-3) multiplied by 2 due to the chain rule.
• Can someone point me to a video on partial fractions? The step from x-5/((2x-3)(x-1)) to A/(2x-3) + B/(x-1) was not entirely clear to me.
• Hi isn't 2 over anything to the minus one integrated, not 2ln? 2ln multiplied by -7/2 is not -7/2ln it would be -7ln? I'm so confused, have I been doing it wrong all this time.

Edit: I have become aware of my mistake I was doing a lot of it in my head and I forgot the du= 2dx would remove the 2 on top of the ln making it 1/(x-3) and so it would be -7/2 multiplied by 1ln and not 2ln. I'm leaving my comment here in hopes that it will help those who have made the same mistake.
• Why does Sal write the answer in terms of the absolute value of the expression? Why not just ln(2x-3), etc?
• I think I've spotted a little mistake at : When you simplify the fraction of -7/(2x-3), you take out -7/2.
Shouldn't the fraction become 1/(x-3) instead of 2/(2x-3) or am I missing a step in this calculation?
(1 vote)
• You're missing a step.
The reason it becomes 2 is because you are putting the derivative of (2x-3) as the numerator.
• Can't you simply simplify and use u-substitution?
(1 vote)
• What do you mean exactly? you could, for example, divide the expression and get the "quotient plus remainder over denominator" form. In fact, the quotient is quite easy to integrate (not even u-substitution is required) but integrating the remainder divided by the denominator requires you to integrate a product of two function...to the minus one power! Unfortunately, i'm not able to do that and it's at least as complex as the problem we started with. Hope that this makes sense
• a question from part of the practice problem:
why
ln|x-1|=ln|2x-2|?
(1 vote)
• They aren't the same. ln|2x-2| equals ln(2) + ln|x-1|. So, there's an additional ln(2) term there.

Considering this is integration, you could condense the ln(2) into the constant of integration at the end. So, ln|x-1| + c and ln|2x-2| + c would be the same (though the value of c in the first expression would actually be (ln(2) + c)). However, in general, the expressions aren't the same.
• I think there is a mistake around . Shouldn’t the equation be A + 3B = -5? Because you have x-5.