Integration by Parts

[cem348]

I've attempted this problem several times and still cannot come to the correct answer. Any help would be greatly appreciated!

Check that integral (x-1) raised to the k dx = 1/(k+1) times (x-1)raised to the k+1 +ct. Use the above formula and integration by parts to find the integral x(x-1) raised to the seventh dx

I was trying to use x-1 for my u term. I keep getting certain integration by parts problems incorrect and I get confused with my book. Is there anywhere that explains a step by step process clearly? Thanks!

 

[soroban]

Hello, cem348!

Do you really understand "Integration by Parts"?

Check that: \(\displaystyle \L\: \int (x\,-\,1)^k\,dx\;=\;\frac{1}{k+1}(x\,-\,1)^{k+1}\,+\,C\)
Use the above formula and integration by parts to find: \(\displaystyle \L\:\int x(x\,-\,1)^7\,dx\)

To check the formula: let \(\displaystyle u \,=\,x-1\;\;\Rightarrow\;\;du\,=\,dx\)

Substitute: \(\displaystyle \L\:\int u^k\,du \;=\;\frac{1}{k+1}u^{k+1}\,+\,C\)

Back-substitute: \(\displaystyle \L\:\frac{1}{k+1}(x\,-\,1)^{k+1}\,+\,C\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Integrate: \(\displaystyle \L\:\int x(x\,-\,1)^7\,dx\)

By parts: \(\displaystyle \L\:\begin{array}{ccc}u\,=\,x & \;\; & dv\,=\,(x-1)^7\,dx \\
\downarrow & & \downarrow \\
du\,=\,dx & \;\; & v \,=\,\frac{1}{8}(x-1)^8 \end{array}\)


And we have: \(\displaystyle \L\:\underbrace{x}_{\text{u}}\,\cdot\,\underbrace{\frac{1}{8}(x-1)^8}_{\text{v}} \,-\,\int\underbrace{\frac{1}{8}(x-1)^8}_{\text{v}}\cdot\,\underbrace{dx}_{\text{du}}\)


. . Then: \(\displaystyle \L\:\frac{1}{8}x(x-1)^8 \,-\,\frac{1}{8}\int(x-1)^8\,dx\)


Can you finish it now?