Integration of linear function to a power: $f\left(ax+b\right)$f(ax+b)
In previous lessons we met the following rules for integration:
| Function $f\left(x\right)$f(x) | Integral $\int f\left(x\right)dx$∫f(x)dx |
|---|---|
| $e^{ax+b}$eax+b | $\frac{1}{a}e^{ax+b}+c$1aeax+b+c |
| $\cos\left(ax+b\right)$cos(ax+b) | $\frac{1}{a}\sin\left(ax+b\right)+c$1asin(ax+b)+c |
| $\sin\left(ax+b\right)$sin(ax+b) | $-\frac{1}{a}\cos\left(ax+b\right)+c$−1acos(ax+b)+c |
In each case we had function composed with a linear function, $ax+b$ax+b. When differentiating such a function we would use the chain rule and multiply the derivative of the function as a whole by the derivative of the inside function (in this case $a$a). To reverse the process we must divide the integral by the constant $a$a.
A form of this integral we have not yet looked at is where $f\left(x\right)=\left(ax+b\right)^n$f(x)=(ax+b)n. That is, where we have a linear function raised to a power.
Recall that to differentiate a linear function raised to a power we would use the chain rule and take follow the following steps:
- Multiply by the index
- Reduce the index by one
- Multiply by the derivative of the function inside the brackets
For example:
| $\frac{d}{dx}\left(2x+5\right)^3$ddx(2x+5)3 | $=$= | $3\times\left(2x+5\right)^{3-1}\times2$3×(2x+5)3−1×2 |
| $=$= | $6(2x+5)^2$6(2x+5)2 |
We have established that integration is the reverse of differentiation. Thus, reversing the process above we obtain the following steps to integrate a linear function raised to a power:
- Divide by the derivative of the function inside the brackets
- Increase the index by one
- Divide by the new index
For example:
| $\int\left(2x+5\right)^2dx$∫(2x+5)2dx | $=$= | $\frac{\left(2x+5\right)^{2+1}}{2\times3}+c$(2x+5)2+12×3+c |
| $=$= | $\frac{1}{6}\left(2x+5\right)^3+c$16(2x+5)3+c |
The following rule can be used to integrate linear functions raised to a power:
$\int\ \left(ax+b\right)^ndx=\frac{\left(ax+b\right)^{n+1}}{a\left(n+1\right)}+c$∫ (ax+b)ndx=(ax+b)n+1a(n+1)+c, $a\ne0$a≠0 and $n\ne-1$n≠−1.
Worked examples
Example 1
Find $\int(2x+1)^3dx$∫(2x+1)3dx.
Think: This is a linear function raised to a power, so we can use the rule above to integrate this expression.
Do: We increase the power by one, divide by the new power and the derivative of the linear function inside the brackets.
| $\int(2x+1)^3dx$∫(2x+1)3dx | $=$= | $\frac{(2x+1)^4}{4\times2}+c$(2x+1)44×2+c |
| $=$= | $\frac{(2x+1)^4}{8}+c$(2x+1)48+c, where $c$c is a constant |
Example 2
Find $\int\frac{2}{(5-x)^4}dx$∫2(5−x)4dx.
Think: There is a linear function raised to a power in the denominator of this expression so we can't use the rule developed above yet. However, if we rewrite this expression using negative indices we can get it into a form that we can integrate.
Do:
| $\int\frac{2}{\left(5-x\right)^4}dx$∫2(5−x)4dx | $=$= | $\int2\left(5-x\right)^{-4}dx$∫2(5−x)−4dx |
Rewrite the expression using negative powers |
| $=$= | $\frac{2\left(5-x\right)^{-4+1}}{-1\times-3}+c$2(5−x)−4+1−1×−3+c |
Add one to the power, then divide by the new power and derivative of the linear function |
|
| $=$= | $\frac{2\left(5-x\right)^{-3}}{3}+c$2(5−x)−33+c |
Simplify |
|
| $=$= | $\frac{2}{3\left(5-x\right)^3}+c$23(5−x)3+c, where $c$c is a constant |
Rewrite with positive powers |
Practice questions
Question 1
Find $\int\frac{4}{\left(x-5\right)^3}dx$∫4(x−5)3dx, with $C$C as the constant of integration.
Question 2
Find $\int\sqrt{7-4x}dx$∫√7−4xdx, with $C$C as the constant of integration.
Question 3
Find $\int\left(5+\left(4x+1\right)^3\right)dx$∫(5+(4x+1)3)dx, with $C$C as the constant of integration.
In each case above we were looking at functions composed with a linear function. What about integrating functions that are composed with more complex functions, such as $\left(x^4+5x\right)^2$(x4+5x)2, $\left(3x^2-2x+1\right)^4$(3x2−2x+1)4, $xe^{3x^2}$xe3x2, or $x^2\sin(4x^3+5)$x2sin(4x3+5)? In the first case we could expand the brackets before integrating, but this strategy is limited to low powers. For instance, we can see that to expand the second case would be very tedious, and furthermore this strategy is of no use in the other two cases as they are not just power functions.
While there are more advanced techniques to integrate such functions, we will approach these by differentiating a related function and looking to manipulate the integral into a form that uses the property that differentiation and integration are reverse operations. That is, if we have a function $y=f\left(x\right)$y=f(x), then:
$\int\frac{dy}{dx}dx=f(x)+c$∫dydxdx=f(x)+c, where $c$c is a constant
Integrating functions in the form $y=kf'(x)[f(x)]^n$y=kf′(x)[f(x)]n
Consider the following
| Function | Derivative using chain rule | Using integration as reverse of differentiation |
Note |
|---|---|---|---|
| $y=(x^2+2)^3$y=(x2+2)3 | $y'=3(x^2+2)^2\times2x$y′=3(x2+2)2×2x $=6x(x^2+2)^2$=6x(x2+2)2 |
$\int6x(x^2+2)^2dx$∫6x(x2+2)2dx $=(x^2+2)^3+c$=(x2+2)3+c |
If $f(x)$f(x) is $(x^2+2)$(x2+2) then $f'(x)$f′(x) is $2x$2x. The function being integrated is in the form $kf'(x)[f(x)]^n$kf′(x)[f(x)]n where $k=3$k=3, $n=2$n=2 |
| $y=(2x^3+5x)^4$y=(2x3+5x)4 | $y'=4(2x^3+5x)^3\times(6x^2+5)$y′=4(2x3+5x)3×(6x2+5) $=4(6x^2+5)(2x^3+5x)^3$=4(6x2+5)(2x3+5x)3 |
$\int4(6x^2+5)(2x^3+5x)^3dx$∫4(6x2+5)(2x3+5x)3dx $=(2x^3+5x)^4+c$=(2x3+5x)4+c |
If $f(x)$f(x) is $(2x^3+5x)$(2x3+5x) then $f'(x)$f′(x) is $6x^2+5$6x2+5. |
| $y=(e^x+3)^7$y=(ex+3)7 | $y'=7(e^x+3)^6\times e^x$y′=7(ex+3)6×ex $=7e^x(e^x+3)^6$=7ex(ex+3)6 |
$\int7e^x(e^x+3)^6dx$∫7ex(ex+3)6dx $=(e^x+3)^7+c$=(ex+3)7+c |
If $f(x)$f(x) is $(e^x+3)$(ex+3) then $f'(x)$f′(x) is $e^x$ex. |
| $y=\sin^4x$y=sin4x $=(\sin x)^4$=(sinx)4 |
$y'=4\sin^3x\times\cos x$y′=4sin3x×cosx $=4\cos x\sin^3x$=4cosxsin3x |
$\int4\cos x\sin^3xdx$∫4cosxsin3xdx $=\sin^4x+c$=sin4x+c |
If $f(x)$f(x) is $(\sin x)$(sinx) then $f'(x)$f′(x) is $\cos x$cosx. The function being integrated is in the form $kf'(x)[f(x)]^n$kf′(x)[f(x)]n where $k=4$k=4, $n=3$n=3 |
In each of the cases above, in the third column the function being integrated was in the form of $kf'(x)[f(x)]^n$kf′(x)[f(x)]n with the integral being $k\times\frac{f(x)^{n+1}}{n+1}$k×f(x)n+1n+1. It's like using the chain rule in reverse! When integrating more complex functions we can look for this type of problem where we have the derivative of a bracket multiplied by the bracket to a higher power and either use a "guess and check" method or a rearranging method to solve the problem.
$\int kf'(x)[f(x)]^ndx=k\times\frac{f(x)^{n+1}}{n+1}+c$∫kf′(x)[f(x)]ndx=k×f(x)n+1n+1+c, $n\ne-1$n≠−1.
Worked examples
Example 3
Determine $\int32x\left(x^2+10\right)^3dx$∫32x(x2+10)3dx.
Method 1: Guess and check
Think: Remember that this question means "this is the derivative, what is the original function?".
Since $32x$32x is a multiple of the derivative of $x^2+10$x2+10 the function is in the form of $kf'(x)[f(x)]^n$kf′(x)[f(x)]n and we can use the chain rule in reverse. We make an educated guess and add one to the power of the bracket.
Do: Guess $y=(x^2+10)^4$y=(x2+10)4
Check by differentiating using chain rule:
$y'=4(x^2+10)^3\times2x=8x(x^2+10)^3$y′=4(x2+10)3×2x=8x(x2+10)3
Compare this to the original question - we have $8x$8x but we need $32x$32x, therefore we must need to multiply our "guess" by $4$4.
Therefore$\int32x\left(x^2+10\right)^3dx=4(x^2+10)^4+c$∫32x(x2+10)3dx=4(x2+10)4+c, where $c$c is a constant.
Method 2: Rearranging
Think: We write the question in the form $k\int f'(x)[f(x)]^ndx$k∫f′(x)[f(x)]ndx then the integral will be $k\times\frac{f(x)^{n+1}}{n+1}+c$k×f(x)n+1n+1+c
Do:
| $\int32x\left(x^2+10\right)^3dx$∫32x(x2+10)3dx | $=$= | $\int16\times2x\left(x^2+10\right)^3dx$∫16×2x(x2+10)3dx |
Rewrite $32$32 as a multiple of $2$2 |
| $=$= | $16\int2x\left(x^2+10\right)^3dx$16∫2x(x2+10)3dx |
Now the function is in the form $k\int f'(x)[f(x)]^ndx$k∫f′(x)[f(x)]ndx |
|
| $=$= | $16\times\frac{(x^2+10)^4}{4}+c$16×(x2+10)44+c |
Add one to the power of the bracket and divide by that number |
|
| $=$= | $4\left(x^2+10\right)^4+c$4(x2+10)4+c, where $c$c is a constant |
Simplify |
example 4
Determine $\int12\sin x\cos^5xdx$∫12sinxcos5xdx.
Method 1: Guess and check
Think: $\cos^5x$cos5x is the same as $(\cos x)^5$(cosx)5
Since $12\sin x$12sinx is a multiple of the derivative of $\cos x$cosx the function is in the form of $kf'(x)[f(x)]^n$kf′(x)[f(x)]n and we can use the chain rule in reverse. We make an educated guess and add one to the power of the bracket.
Do: Guess $y=(\cos x)^6$y=(cosx)6
Check by differentiating using chain rule:
$y'=6(\cos x)^5\times-\sin x=-6\sin x\cos^5x$y′=6(cosx)5×−sinx=−6sinxcos5x
Compare this to the original question - we have $-6$−6 but we need $12$12, therefore we must need to multiply our "guess" by $-2$−2.
Therefore$\int12\sin x\cos^5xdx=-2\cos^6x+c$∫12sinxcos5xdx=−2cos6x+c, where $c$c is a constant.
Method 2: Rearranging
Think: We write the question in the form $k\int f'(x)[f(x)]^ndx$k∫f′(x)[f(x)]ndx then the integral will be $k\times\frac{f(x)^{n+1}}{n+1}+c$k×f(x)n+1n+1+c
Do:
| $\int12\sin x\cos^5xdx$∫12sinxcos5xdx | $=$= | $-12\int-\sin x\cos^5xdx$−12∫−sinxcos5xdx |
The derivative of $\cos x$cosx is $-\sin x$−sinx so we need to multiply $\sin x$sinx by $-1$−1 and hence $12$12 by $-1$−1. Now the function is in the form $k\int f'(x)[f(x)]^ndx$k∫f′(x)[f(x)]ndx |
| $=$= | $-12\times\frac{(\cos x)^6}{6}+c$−12×(cosx)66+c |
Add one to the power of the bracket and divide by that number |
|
| $=$= | $-2\cos^6x+c$−2cos6x+c, where $c$c is a constant |
Simplify |
Practice questions
Question 4
Consider the function $y=e^{x^3}$y=ex3.
Find the derivative $\frac{dy}{dx}$dydx.
Hence find the value of $\int3x^2e^{x^3}dx$∫3x2ex3dx.
You may use $C$C to represent the constant of integration.
Question 5
Consider the function $y=\sin\left(x^5\right)$y=sin(x5).
Find the derivative $\frac{dy}{dx}$dydx.
Hence find the value of $\int x^4\cos\left(x^5\right)dx$∫x4cos(x5)dx.
You may use $C$C to represent the constant of integration.
Using the strategy of integration as the reverse of differentiation
Worked example
Example 5
Given $f(x)=xe^x$f(x)=xex.
(a) Find $f'(x)$f′(x).
Think: Here the function is of the form of the product of two functions, so we will use the product rule.
Do:
| Let | $u=x$u=x | then | $u'=1$u′=1 |
| and | $v=e^x$v=ex | then | $v'=e^x$v′=ex |
| $f'\left(x\right)$f′(x) | $=$= | $uv'+vu'$uv′+vu′ |
Write out the product rule |
| $=$= | $x\times e^x+e^x\times1$x×ex+ex×1 |
Make the appropriate substitutions |
|
| $=$= | $xe^x+e^x$xex+ex |
Simplify |
(b) Hence, determine the integral $\int xe^xdx$∫xexdx.
Think: Since integration is the reverse of differentiation, part (a) tells us that $\int xe^x+e^xdx=xe^x+c$∫xex+exdx=xex+c, for some constant $c$c.
We have been asked to determine the integral of the first term of the function inside the integral sign so we will use the property $\int f\left(x\right)+g\left(x\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx$∫f(x)+g(x)dx=∫f(x)dx+∫g(x)dx to split the integral and then rearrange to find the part we require.
Do:
| $\int xe^x+e^xdx$∫xex+exdx | $=$= | $xe^x+c$xex+c |
From part (a) |
| $\int xe^xdx+\int e^xdx$∫xexdx+∫exdx | $=$= | $xe^x+c$xex+c |
Using the property $\int f\left(x\right)+g\left(x\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx$∫f(x)+g(x)dx=∫f(x)dx+∫g(x)dx |
| $\int xe^xdx+e^x$∫xexdx+ex | $=$= | $xe^x+c$xex+c |
Evaluate $\int e^xdx$∫exdx |
| $\therefore\int xe^xdx$∴∫xexdx | $=$= | $xe^x-e^x+c$xex−ex+c, where $c$c is a constant |
Rearrange |
Practice questions
Question 6
Consider the following.
Determine $\frac{d}{dx}\left(x\sin x\right)$ddx(xsinx).
Hence find $\int4x\cos xdx$∫4xcosxdx.
You may use $C$C as the constant of integration.