tell me the integration of x power 3 minus 1 whole power1 upon 3 into x power 5 DX
Answers
Answer:
THE METHOD OF INTEGRATION BY POWER SUBSTITUTION
The following problems involve the method of power substitution. It is a method for finding antiderivatives of functions which contain $n$th roots of $ x $ or other expressions. Examples of such expressions are
$ \displaystyle{ 5 \over x^{1/2} + x^{1/3} } $
and
$ \displaystyle{ 3x-7 \over \sqrt[3] { 5x+2 } } $ .
The method of power substitution assumes that you are familiar with the method of ordinary u-substitution and the use of differential notation. Recall that if
$ u = f(x) $ ,
then
$ du = f'(x) \ dx $ .
For example, if
$ u = 3x^2 - 5x + 7 $ ,
then
$ du = (6x-5) \ dx $ .
When using the method of power substitution, we will usually assume that
$ x = f(u) $ ,
so that
$ dx = f'(u) \ du $ .
For example, if
$ x = u^{3} $ ,
then
$ dx = (3u^{2}) \ du $ .
The goal of a power substitution will be to replace $n$th roots of functions, which may not be easily integrated, with integer powers of functions, which are more easily integrated. For example, if we start with the expression
$ x^{1/5} $
and let
$ x = u^{5} $ ,
then
$ x^{1/5}= (u^{5})^{1/5} = u^{5(1/5)} = u^1 = u $
and
$ dx = (5u^{4}) \ du $ .
Thus,
$ \displaystyle{ \int x^{1/5} \, dx } $
could be rewritten as
$ \displaystyle{ \int u(5u^4) \, du } = \displaystyle{ \int 5u^5 \, du } $ .
In general, to eliminate
$ x^{1/n} $ ,
let
$ x = u^{n} $ .
Then
$ x^{1/n}= (u^{n})^{1/n} = u^{n(1/n)} = u^1 = u $
and
$ dx = nu^{n-1} \ du $ .
Recall the following well-known, basic indefinite integral formulas :
1.) $ \displaystyle{ \int x^n \,dx } = \displaystyle{ {x^{n+1} \over n+1 } + C } $ , where $ n \ne -1 $
2.) $ \displaystyle{ \int { 1 \over x } \,dx } = \displaystyle{ \ln \vert x\vert + C } $
3.) $ \displaystyle{ \int { 1 \over x^2 + a^2 } \,dx } = \displaystyle{ {1 \over a} \arctan {x \over a} + C } $
4.) $ \displaystyle{ \int k f(x) \,dx } = k \displaystyle{ \int f(x) \,dx } $ , where $ k $ is a constant
5.) $ \displaystyle{ \int ( f(x) \pm g(x) ) \,dx } = \displaystyle{ \int f(x) \,dx } \pm \displaystyle{ \int g(x) \,dx } $
We can now apply the method of power substitution to an integral like
$ \displaystyle{ \int { 1 \over x^{1/5} + 2 } \, dx } $ .
A COMMON WRONG ANSWER FOR THIS PROBLEM IS
$ \displaystyle{ \int { 1 \over x^{1/5} + 2 } \, dx }
= \ln \vert x^{1/5} + 2 \vert + C $ .
Convince yourself that the answer is wrong by differentiating the right-hand side and showing it is NOT equal to the function on the left-hand side of the above equation. To properly integrate this problem, use the power substitution
$ x = u^{5} $
so that
$ x^{1/5}= (u^{5})^{1/5} = u^{5(1/5)} = u^1 = u $
and
$ dx = 5u^{4} \ du $ .
Now substitute into the original problem, replacing all forms of $ x $, getting
$ \displaystyle{ \int { 1 \over x^{1/5} + 2 } \, dx }
= \displaystyle{ \int { 1 \over u + 2 } \, 5u^{4} \ du } $
$ = \displaystyle{ \int { 5u^{4} \over u + 2 } \, du } $
(Use polynomial division.)
$ = \displaystyle{ \int \Big( 5u^3-10u^2+20u-40+ {80 \over u+2 } \Big) \, du } $
$ = \displaystyle{ \int \Big( 5u^3-10u^2+20u-40+ 80{1 \over u+2 } \Big) \, du } $
$ = \displaystyle{ 5{u^4 \over 4} - 10{u^3 \over 3} + 20{u^2 \over 2} - 40u + 80 \ln \vert u+2\vert } + C $
$ = \displaystyle{ {5 \over 4}(x^{1/5})^4 - {10 \over 3}(x^{1/5})^3 + 10(x^{1/5})^2 - 40x^{1/5} + 80 \ln \vert x^{1/5}+2\vert } + C $
$ = \displaystyle{ {5 \over 4}x^{4/5} - {10 \over 3}x^{3/5} + 10x^{2/5} - 40x^{1/5} + 80 \ln \vert x^{1/5}+2\vert } + C $ .
Most of the following problems are average. A few are challenging. Make careful and precise use of the differential notation $ dx $ and $ du $ and be careful when arithmetically and algebraically simplifying expressions. You will likely need to use the following rules for exponents.
$ (AB)^m = A^m B^m $
$ (A^m)^n = A^{mn} $
$ A^m A^n = A^{m+n} $
$ \displaystyle{ A^m \over A^n } = A^{m-n} $