Dick said:What's wrong with u=x/4+6? Can't you express x+25 in terms of u?
percs said:well then du = dx ... umm how could that be correct?
Dick said:What's wrong with u=x/4+6? Can't you express x+25 in terms of u?
Inferior89 said:If u = x/4 +6 then du is not equal to dx.
percs said:well if u = x/4 +6 then the closest to x+25 is 4u = x+24
percs said:well then du = dx ... umm how could that be correct?
Inferior89 said:The point is to get rid of all the x. You don't need it to have the form c*u where c is some constant.
Dick said:i) du isn't equal to dx and ii) how do you know it couldn't work if you haven't tried it?
percs said:i did it gives me u = x/4 + 6 and du = dx/4
dont know where to go form there
fzero said:You have an algebra mistake in your substitution. u = x/4 + 6 and x = 4u - 24 mean that
(x+25)(x/4+6)^7 dx = ( 4u -24 + 25) (u)^7 ( 4 du)
An integral substitution is a method used in calculus to simplify and solve integrals by substituting a variable or expression with a new one. This is often used when the integrand (the function being integrated) contains a complicated expression that can be simplified with the substitution.
There is no exact rule for when to use an integral substitution, but it is often helpful when the integrand contains a complicated expression that can be simplified with the substitution. It is also helpful when the integrand contains a function and its derivative, or when it contains a function raised to a power.
The steps for performing an integral substitution are as follows: 1) Identify the substitution to be made by looking for a function and its derivative, or a function raised to a power. 2) Substitute the new variable or expression into the integral. 3) Rewrite the integral in terms of the new variable or expression. 4) Solve the integral using traditional integration techniques. 5) Substitute the original variable or expression back into the final solution.
No, not all integrals can be solved using an integral substitution. Some integrals may require other techniques such as integration by parts or partial fractions. It is important to first try an integral substitution, and if it does not work, then try other methods.
Choosing the right substitution can be challenging, but there are a few tips that can help. 1) Look for expressions in the integrand that can be simplified with the substitution. 2) Try to choose a substitution that will eliminate or reduce the number of terms in the integrand. 3) Be aware of common substitutions, such as u-substitution for functions and their derivatives, or trigonometric substitutions for expressions involving trigonometric functions.