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Substitution and parts


The method of integration by substitution derives from the chain rule, or composite function rule:


In an integral of a binomial expression, such as $(ax + b)^n$, it would be possible to first exapnd the expression, and then take the integral with respect to x. However, if the value of $n$ is large, this would become rather tedious and impractical. If the binomial expression is the denominator, it may also be close to impossible to work out the integral with respect to x.

But, do not panic - help is at hand with the substitution method. By replacing the binomial, $ax + b$, by a letter (traditionally $u$), the composite function rule allows us to find the integral of $u$ with respect to $u$ (i.e. $du$), and to ensure this gives the integral with repsect to $x$, we multiply the integral of $u$ by the 'internal' integral of the binomial.

That was a mouthful, so let's do an example:

$∫(3x + 4)^5dx$

First we assign $u = 3x + 4$, and see that ${du}/{dx} = 3$

We substitute these values into the original equation:

$∫(3x + 4)^5dx = ∫(u)^5dx$

Now we have something we can't solve: u to the respect of x. So, we need to change the $dx$ to a $du$, by inserting the ${du}/{dx} = 3$. To make sure the equation balances, we can only multiply by 1, so we rearrange the equation to:${{du}/{dx}}/{3} = 1$.

Now we put this little sleight of hand into our equation and see something we can integrate:

$∫(u)^5dx = ∫{{du}/{dx}}/{3}u^5dx = ∫{u^5}/{3}du = 1/3∫u^5du = 1/3{u^6}/6 + C = 1/{18}{u^6} + C$

And now we substitute the binomial expression back in so it makes sense, since we invented the $u$:

$∫(3x + 4)^5dx = 1/{18}(3x + 4)^6 + C$

Integration by Parts

Integration by parts is the counterpart of the product rule for differentiation:

$${d(u⋅v)}/{dx} = {du}/{dx}⋅v + u⋅{dv}/{dx}$$

Integrating with respect to x results in:

$∫{d(u⋅v)}/{dx} = ∫({du}/{dx}⋅v + u{dv}/{dx})dx = ∫{du}/{dx}⋅vdx + ∫u{dv}/{dx}dx$ ⇒ $uv = ∫{du}/{dx}⋅vdx + ∫u{dv}/{dx}dx$, or:

$$∫u{dv}/{dx}dx = uv - ∫{du}/{dx}vdx$$

where $u$ and $v$ are functions of $x$.


Find the integral of $3xe^x$ with respect to x.

Let $u = 3x$ ⇒ ${du}/{dx} = 3$, or $du = 3dx$.

If ${dv}/{dx} = e^x$, then $v = ∫e^xdx = e^x$.

$∫3xe^xdx = uv - ∫v{du}/{dx}dx = 3xe^x - ∫e^x⋅3dx =$

$= 3xe^x - 3e^x + c = 3e^x(x - 1) + c$

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