Solving integrals

[Math_Junkie]

#1.

I expanded the top and cancelled out the x's. I end up with logarithms involved and I'm not sure how to handle them in order to simplify my answer.

#2.

I know substitution method is involved here.. but I keep running into dead ends.

Thanks in advance.

 

[Deleted member 4993]

#1.

I expanded the top and cancelled out the x's. I end up with logarithms involved and I'm not sure how to handle them in order to simplify my answer.

#2.

I know substitution method is involved here.. but I keep running into dead ends.

Thanks in advance.

Please show us your work - so that we know where to begin to help you.

for the second problem, substitute:

u = x[sup:241lxfuy]16[/sup:241lxfuy] + 9

du/dx = 16 * x[sup:241lxfuy]15[/sup:241lxfuy]

Where is the dead end?

 

[Math_Junkie]

I figured out #2. Thank you.

For #1:

9 (x[sup:2q776qut]3[/sup:2q776qut] - 3x[sup:2q776qut]2[/sup:2q776qut] + 3x - 1)/x[sup:2q776qut]2[/sup:2q776qut])
9 (x - 3 + 3/x - 1/x[sup:2q776qut]2[/sup:2q776qut])
9x - 27 + 27/x - 9/x[sup:2q776qut]2[/sup:2q776qut]

The antiderivative is: 9/2x[sup:2q776qut]2[/sup:2q776qut] - 27x + 27logx + 9x[sup:2q776qut]-1[/sup:2q776qut]

It's when I start plugging in for F(4) - F(1) that I get stuck. I don't know how to simplify to give me an exact answer.

 

[Deleted member 4993]

I figured out #2. Thank you.

For #1:

9 (x[sup:xu82oxyy]3[/sup:xu82oxyy] - 3x[sup:xu82oxyy]2[/sup:xu82oxyy] + 3x - 1)/x[sup:xu82oxyy]2[/sup:xu82oxyy])
9 (x - 3 + 3/x - 1/x[sup:xu82oxyy]2[/sup:xu82oxyy])
9x - 27 + 27/x - 9/x[sup:xu82oxyy]2[/sup:xu82oxyy]

The antiderivative is: 9/2x[sup:xu82oxyy]2[/sup:xu82oxyy] - 27x + 27logx + 9x[sup:xu82oxyy]-1[/sup:xu82oxyy]

It's when I start plugging in for F(4) - F(1) that I get stuck. I don't know how to simplify to give me an exact answer.

Did you simplify the algebraic part?

Also did you use log(1) = 0

 

[Math_Junkie]

Did you simplify the algebraic part?

Also did you use log(1) = 0

I just plugged in for F(4) - F(1).While solving for this I did use log(1) = 0.

The final answer that I ended up with was: -81/4 + 27log4, but it's incorrect.

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ \int_{1}^{4} 9\frac{(x-1)^{3}}{x^{2}} \ dx \ = \ 9\int_{1}^{4}\frac{x^{3}-3x^{2}+3x-1}{x^{2}} dx\)

\(\displaystyle = \ 9\int_{1}^{4}\bigg[x-3+\frac{3}{x}-\frac{1}{x^{2}}\bigg]dx \ = \ 9\bigg[\frac{x^{2}}{2}-3x+3ln|x|+\frac{1}{x}}\bigg]_{1}^{4}\)

\(\displaystyle = \ 9\bigg[8-12+3ln(4)+\frac{1}{4}-\bigg(\frac{1}{2}-3+3ln(1)+1\bigg)\bigg]\)

\(\displaystyle = \ \bigg [-\frac{81}{4}+27ln(4)\bigg] \ = \ 17.1799477502\)

\(\displaystyle Now,\ where \ is \ the \ problem?\)