Integrate y = x ( x^2 +1 )^3 dx?

Question

Use the substitution u=x^2 +1 to integrate x ( x^2 +1 )^3 dx with the bounds (2 and 1).

Answer 1

du/dx= 2x
du=2x dx

y= (x^2+1)^3* 2x/2 dx
y= u^3 du/2
integral y= u^4/(4*2)+c
= u^4/8
u=x^2+1
when x=1 u=2
when x=2 u=5
definite integral
=f(5)-f(2)
= (5^4)/8 - (2^4)/8
=76.125

Answer 2

u = x^2 + 1
u - 1 = x^2
sqrt(u - 1) = |x|

If we take an integral from x = 1 to x = 2, or indeed as you stated, from x = 2 to x = 1, in either case x is always positive so we can drop the absolute value operation. I am going to assume that you intended 1 to be the lower bound. If not, just change the sign on our final answer.

x = sqrt(u - 1)
dx = 1 /(2*sqrt(u - 1)) du

Now we can rewrite our integral as follows, changing our limits of integration to be in terms of u instead of x

∫[from u=2 to u=5] (sqrt(u - 1)*u^3) / (2sqrt(u - 1)) du
= ∫[2:5] (u^3)/2 du
= (1/8)u^4 ]{2:5}

we can now input directly
= (1/8)[5^4 - 2^4]
= 609/8

or return to a form in terms of x
=(1/8)(x^2 + 1)^4]{1:2}
=(1/8)[5^4 - 2^4]
= 609/8

--charlie

Answer 3

integrate y= x(x^2 + 1)^3 dx
you must use subsitution by u
u = x^2 + 1
differential u => du = 2x dx
but when we want to integrate we just use dx so we must appear dx by...
dx = du/2x
so we can solve the integrate y = x (u)^3 du/2x
we can eliminate x and the change is 1/2
so... 1/2 . integrate y = (u)^3
1/2. 1/4 (u)^4
1/8 (u)^4
u = x^2 +1
so the result is...
1/8 (x^2 +1)^4

Answer 4

Integrate y = x ( x^2 +1 )^3 dx
u=x^2 +1
du/dx=2x
dx=du/2x
int x*( x^2 +1 )^3 dx
= int(x*u^3(1/2x))du
=(1/2)*int(u)^3du
=(1/2)*(u)^4/4+C
=u^4/8+C
but,u=x^2 +1
hence,
int(x ( x^2 +1 )^3) dx
=(x^2+1)^4/8 between the
bounds x=2 to x=1
=5^4/8-2^4/8
=(625-16)/8
=76.125

i hope that this helps

Answer 5

Int[x ( x^2 +1 )^3 dx]

Making the sub gives

Int[x*u^3 dx]

If u=x^2 +1 then du/dx=2x so du/2x = dx

Int[0.5(u^3) du]

= (1/8)u^4

Now the limits where x1=1 and x2=2 but we are working with u

u=x^2 +1

u1 = 2 and u2 = 5

(1/8)(5)^4 - (1/8)(2)^4 = 76.12500

Answer 6

1. you find the derivative of the equation you let u equal.
ie. du/dx of (x^2+1) which equals 2x.
2. you re-write the intergral from 1 to 2 (or 2 to 1) depending on which one you're after - i wasn't sure what you ment by, "with the bounds 2 and 1" as you should have been given a starting point to the end point. ie from 2 to 1 or from 1 to 2 (which ever one its from, then that number is on the bottom part of the intergral sign.)
3.as your intergral is now the intergral of xu^3 dx ( as u=(x^2)+1 )
- now think of how can you get u on its own and dx to du as you can't intergrate u with a dx it has to be intergrated with a du.
4. you do this by using your du/dx = 2x equation from part 1. rearrange it so you have du/2x = dx
5. now go back to your intergral equation and sub dx for dx= du/2x
6. so now you have the intergral of xu^3 du/2x ( don't forget your terminals on the intergral sign. - the 2 and the1 )
7. cancel out the x's
8. now you have intergral of u^3/2 du
9. intergrate to get [u^4/8] ( with your terminals)
10. now remeber u = x^2 + 1? - sub that back into ur new equation - ie so you get now [ ((x^2 + 1)^4)/8 ]
11. sub in ur terminals into the equation - top terminal number minus the bottom terminal number to get the answer

if ur terminals are 1 to 2 then it should be 609/8
and if it's the other way around its just the negative value. -609/8

hope this helps and makes sence to you!

Answer 7

u=x²+1» u' =2x »» x = u'/2
x= 2, u = 5
x = 1, u = 2

So Int [x(x²+1)³] = Int 1/2 [ u³ u' ] = 1/8 u^4 , u ranging from 2 to 5 (or from 5 to 2, dependig of what extreme comes first, I assumed 2 is before 5)

So Int [x(x²+1)³] = 1/8 (5^4 - 2^4)
Pls do yourselve .

Answer 8

I = ∫ x.(x² + 1)³ dx
Let u = x² + 1
du/dx = 2x and thus du/2 = x.dx
I = (1/2).∫ u³.du
I = (1/8).u^(4) between lims. of 1 and 2
I = (1/8).(x² + 1)^(4) between lims 1 and 2
I = (1/8)[ 5^(4) - 2^(4) ]
I = (1/8) x 609 = 76.1

Answer 9

Y=x(x^2+1)^3
Integrate from 2 to 1
x(x^2+1)^4/8x
[2(2^2+1)^4/8(2)] –[1(1^2+1)^4/8(1)]
=78.125-0.125
=78

Answer 10

u=x^2+1
du=2xdx
at x=1==>u=2
at x=2==>u=5
i=int(u^3 du/2) from u=2 to u=5
=0.5*int(u^3du) from u=2 to u=5
=0.5/4*u^4 from u=2 to u=5
=0.125(5^4-2^4)=0.5(625-16)
=76.125