I think I see where you are coming from now. The problem is the 13x in the denominator.

Your Short cut way:

int[ (2x+3)^12]dx

= (1/13) (2x+3)^13 / 2 ---> divide by the inner derivative of 2x+3 which is 2 to undo
the chain rule. Not x as you were doing

= (1/26) (2x+3)^13 + c

Check by taking the derivative:

d/dx [ (1/26)(2x+3)^13]

= (1/26) (13) [(2x+3)^12)] (2) --------> times by 2 as the result of the chain rule.

= (26/26)(2x+3)^12

= (2x+3)^12

I should point out that this little short cut only works when the inner derivative is a constant - in this case 2. If you are integrating ( 2x^2 + 3x)^13 then you are SOL.

@_ sunflower Yes, if the inner function is linear.

But your short cut can be used in other special cases. For example if you wanted to integrate: 2x^2*(x^3+5)^1/2. You will notice that the inner derivative is 3x^2 and you conveniently have a factor of 2x^2 out in front so just using the power rule on

(x^3 + 5)^ (1/2) to give (x^3+5)^(3/2) your outside factor be out by a constant when you use the chain rule. So take the derivative of your first attempt to see what that constant is. Don't divide by 3x^2.

So you can see that you have a factor 9/2 out in front instead of the desired 2 in the original integral so to change (9/2) into 2 just multiple by (4/9)

Finally: integral of 2x^2 *(x^3+5)^1/2 = (4/9)*(x^3+5)^3/2 + c

Remember, differentiating your final answer is the ultimate test.

Integrating (x^2+3) is just an example of why I came up with the idea of integrating tan^2x using the second method from the first doc.

So my school taught me that, when integrating complex expressions for example (2x+3)^12. You don't have to expand it arduously, but just plus one to the power and divide it to itself, also differentiate the expression inside the bracket and then divide it to itself again. [(2x+3)^13]/13x

I think you are referring only to the file that I replied to the first response. Could you please check the first document that I attached in the question?

I think I see where you are coming from now. The problem is the 13x in the denominator.

Your Short cut way:int[ (2x+3)^12]dx

= (1/13) (2x+3)^13 /

2---> divide by the inner derivative of 2x+3 which is 2 to undo the chain rule. Not x as you were doing= (1/26) (2x+3)^13 + cCheck by taking the derivative:d/dx [ (1/26)(2x+3)^13]

= (1/26) (13) [(2x+3)^12)] (2) --------> times by 2 as the result of the chain rule.

= (26/26)(2x+3)^12

= (2x+3)^12

I should point out that this little short cut only works when the

inner derivative is a constant- in this case 2. If you are integrating ( 2x^2 + 3x)^13 then you are SOL.Thank you for your kind and amazing reply again.

Yeah I made a careless mistake there hahahahaha.

So the same idea goes to the integration of tan^2x? The inner derivative must be constant when using the short cut?

@_ sunflower Yes, if the inner function is linear.

But your short cut can be used in other special cases. For example if you wanted to integrate:

2x^2*(x^3+5)^1/2. You will notice that the inner derivative is 3x^2 and you conveniently have a factor of 2x^2 out in front so just using the power rule on(x^3 + 5)^ (1/2) to give

(x^3+5)^(3/2)your outside factor be out by a constant when you use the chain rule. So take the derivative of your first attempt to see what that constant is. Don't divide by 3x^2.d/dx[ (x^3 + 5)^3/2]= (3/2)*(x^3 + 5)^1/2 * 3x^2 =(9/2) x^2*( x^3 + 5)^1/2So you can see that you have a factor 9/2 out in front instead of the desired 2 in the original integral so to change (9/2) into 2 just multiple by (4/9)

Finally:

integral of 2x^2 *(x^3+5)^1/2 = (4/9)*(x^3+5)^3/2 + cRemember, differentiating your final answer is the ultimate test.

d/dx[ (4/9) *(x^3+5)^3/2] = (4/9)*(3/2)*(x^3+5)^1/2 * (3x^2) =

2x^2(x^+5)^(1/2)This short cut only works if the outside factor is a constant*inner derivative.

@Ian Fowler Thank you sooo sooo much for you kindness!! Now I understand and know that I should be careful with the constant!

Since tan(x) whole square is not equal to {tan(tanx)}^2 thats why the second one is absolutely wrong.

Since tan(x) whole square is not equal to {tan(tanx)}^2 thats why the second one is absolutely wrong.

Since tan(x) whole square is not equal to {tan(tanx)}^2 thats why the second one is absolutely wrong.

this is the simpler version of the integral method that my school taught me. And I applied it just the same to tan^2x

Isn't tan^2x tanx X tanx? which is equal to (tanx)^2 ?

Integrating (x^2+3) is just an example of why I came up with the idea of integrating tan^2x using the second method from the first doc.

So my school taught me that, when integrating complex expressions for example (2x+3)^12. You don't have to expand it arduously, but just plus one to the power and divide it to itself, also differentiate the expression inside the bracket and then divide it to itself again. [(2x+3)^13]/13x

I think you are referring only to the file that I replied to the first response. Could you please check the first document that I attached in the question?