Using U-Substitution to Find an Integral

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So I was messing around with some basic u-sub calculus and came across this problem. Any help would be greatly appreciated!

Homework Statement



\int \frac{3x^{2}}{x^{2}-7} dx

Homework Equations


I'm using u-substitution, as stated at the beginning.

The Attempt at a Solution


I see that the u could be the denominator and the numerator is the derivative, or du.

\int \frac{du}{u}

\int \frac{1}{u} du

ln|u|+C

ln|x^{2}|+C

I'm pretty sure this is the correct answer, but I decided to integrate it with the boundaries x=1 and x=4.5.

[ln|x^{2}-7|]^{4.5}_{1}

ln|13.25| - ln|-6|
ln(13.25) - ln(6)
2.5839-1.7917
.7922

So this is my final answer, but it just doesn't seem right. I graphed the function and saw it was under the x-axis. Is this the problem?
 
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lch7 said:
So I was messing around with some basic u-sub calculus and came across this problem. Any help would be greatly appreciated!

Homework Statement



\int \frac{3x^{2}}{x^{2}-7} dx

Homework Equations


I'm using u-substitution, as stated at the beginning.

The Attempt at a Solution


I see that the u could be the denominator and the numerator is the derivative, or du.

But if ##u=x^2-7## then ##du = 2xdx## and that is not what you have in the numerator. Try one long division step and see what to do next.
 
Following what Lc suggested, here's a hint that will make it easier : ##x^2 - 7 = (x + \sqrt{7})(x - \sqrt{7})##.
 
Oh man I feel like an idiot! Thankyou all!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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