Antiderivative of a polynomial

Euler

Junior Member
Joined
Jun 28, 2005
Messages
102
I am trying to find the anti-derivative to
\(\displaystyle (4x+5)^7\)

The book tells me it is \(\displaystyle \frac{1}{32}(4x+5)^8+C.\)

I don't understand how they arrive at this answer. I took a rather lengthy way, expanding the entire polynomial using Pascal's triangle, and then taking the antiderivative of each individual term. They seem to do it a much quicker way. How?
 

Euler

Junior Member
Joined
Jun 28, 2005
Messages
102
Wait a second, did they just make \(\displaystyle (4x+5) = \frac{1}{4}(x+4/5)\) and then divide the fraction by 8 (effectively taking the antiderivative)?

This would give me \(\displaystyle \frac{1}{32}(x+\frac{4}{5})^8\) and the obvious + C (because I am forming a general equation.) How do they get it back to (4x+5)?
 

Unco

Senior Member
Joined
Jul 21, 2005
Messages
1,134
Euler! If you can't see it straight away, no problem. Use a u-substiution!

Let u = 4x + 5
du/dx = 4 --> dx = du/4

\(\displaystyle \int { (4x+5)^7 \, dx} = \int { u^7 \, \frac{du}{4}}\)

\(\displaystyle = \int{ \frac{1}{4} u^7 \, du} = \frac{1}{4} \cdot \frac{1}{8}u^8 + C = \frac{1}{32}u^8 + C\)

Subsitute back for u
\(\displaystyle \frac{1}{32}(4x+5)^8 + C\)
 
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