integral problem using the substitution rule?

Question

indefinite integral of sqrt(1+x^2) x^5 dx

my professor worked this problem out for us in class, however i don't understand how he got certain parts. i find the substitution rule confusing. please explain the steps in solving this problem.

Answer 1

The idea behind the substitution rule is that if do not know how to find the integral in terms of x, you can write it in terms of another variable u and then find the integral in terms of u, then resubstitute the original expression in terms of x. For this instance, the logical choice of u is 1+x², in which case du=2x dx, i.e. dx = du/(2x). So making the substitution, we have:

∫√u x^5/(2x) du

Of course, we still have a variable x in an integral with respect to u. Not good. Fortunately, after simplification we have:

1/2 ∫√u x^4 du

Now, recall that u=1+x², so x²=u-1, and thus x^4 = (u-1)². So we have:

1/2 ∫√u (u-1)² du

Expanding the (u-1)²:

1/2 ∫√u (u²-2u+1) du

Now distributing the √u:

1/2 ∫u^(5/2) - 2u^(3/2) + √u du

And now we can integrate this using the power rule:

1/7 u^(7/2) - 2/5 u^(5/2) + 1/3 u^(3/2) + C

And now substituting u=1+x²:

1/7 (1+x²)^(7/2) - 2/5 (1+x²)^(5/2) + 1/3 (1+x²)^(3/2) + C

And now we are done.

Answer 2

step 1. Let u = 1 + x^2. That means du/dx = 2x ==> dx = du/2x

Step 2. substitute 1+x^2 for u and du/2x for dx. The integral then becomes:

sqrt(u)* x^5 * du/2x, which simplifies to (1/2) x^4 * (sqrt(u))

Step 3. This is the tricky part. If u = 1 + x^2, then x^2 = u-1. This means that the x^4 piece can also be written in terms of u because x^4 = (x^2)^2 = (u-1)^2

Step 4. You are now left with:

(1/2) * sqrt(u) * (u-1)^2 du

Treat sqrt (u) as u^(1/2) and simply FOIL and distribute.

Step 5: integrate using the power rule, and after you're done, replace u with 1+ x^2

Answer 3

Let u = 1+x², then
du = 2x dx --> .5du = x dx
AND
x = sqrt(u-1)
thus
x^4 = (u-1)²
Let's put it all together:

∫sqrt(1+x²)*x^5 dx
= ∫sqrt(1+x²)*x^4 * x dx
= ∫sqrt u * (u-1)² * .5 du
=.5∫u^.5 (u² - 2u + 1) du
= .5 ∫u^2.5 - 2u^1.5 + u^.5 du
= .5 (2/7 u^3.5 - 4/5 u^2.5 + 2/3 u^1/5) + C
= 1/7 u^3/5 - 2/5 u^2.5 + 1/3 u^1/5 + C

Substitute back in u=1+x² and that's the whole enchilada!

Answer 4

I think you have an integral by parts problem in which you substitute.
It goes Z UdV = UV - Z VdU
"Z" is a cheap integral sign, and Uand V are functions of x that you will subsitute for. But first, a very sneaky trick. We let x^2= r and dr=2xdx
Now our problem is transformed to one in "r" and looks like
Z r^2 sqrt(1+r) /2 dr
That's a lot easier to integrate by parts,

Answer 5

6?(3) ? [ dx ] / [ a million / 36( a million + (x^2/36) ] 0 6?(3) ? (a million/36) [ dx ] / [ a million / ( a million + (x/6)^2 ] 0 u = x/6 = (a million/6)x du = (a million/6) dx ===> dx = 6 du 6?(3) ? (a million/36) [ 6du ] / [ a million / ( a million + (u)^2 ] 0 6?(3) ? (a million/6) [ du ] / [ a million / ( a million + (u)^2 ] 0 . .. . . .. . . .. . . .. . 6?(3) (a million/6) * tan^-a million( u ) ] = (a million/6) * [ tan^-a million( x/6 ) ] = (a million/6) * [ tan^-a million( 6?(3)/6 ) - tan^-a million( 0/6 ) ] . .. . . .. . . .. . . . . .0 (a million/6) * [ tan^-a million( ?(3) ) - tan^-a million( 0 ) ] (a million/6) * [ ?/3 - 0 ] = ?/18