TRICKY integral! looks easy but is hard please help?

Question

what is the integral of [1/(1+x^3)] dx from 0 to 3
please show your work so I can understand
thanks soooo much

Answer 1

It is indeed nasty:

The indefiniate integral is:
1/3 (1/2 log((1 + x) ^3/(1+x^3)) + arctan((-1 + 2 x) / sqrt(3)) sqrt(3) ))

You can plug in 0 and 3 to see what you get.

I used a computer, so no step-by-step. Sorry

Answer 2

let I=integral [1/1+x^3]dx frm 0to3

put x=tanA
so dx/dA=sec^2A
dx=sec^2A.dA
now
I=integral[sec^2A/tan^3A+1]dA frm 0to tan inverse3

the limits get changed because if u put x =0 in x =tanA then A=0
similarly on putting x=3 we get A=tan inverse3

now i guess u can solve them urself

hope u got ur answer :-))

Answer 3

Since this sounds like school, I'll only give you a hint:

try factoring 1 + x^3

Your integral will be a lot easier when you do.

Answer 4

very easy.....

just make a substitution....

here we go!!!!

u=1+X^3

du = 1+3X^2 dX

du/3 = X^2 dX


so..... Int defines from 0 to 3 1/u *du/3

remembre this.... int 1/X = ln x

so..... 1/3 * int def 0 to 3 1/u du note this ( you can take a 1/3 from integral)

this is equal to...... 1/3 ( ln u) from 0 to 3

and thats all.... just substitute u because you know u = 1+X^3

good luck!!!! finish the rest.... I think any teacher will accept this answer to this point , but you can keep on resolving the problem

Answer 5

I hate to do integration online, i just cant be bothered to read the equations in this strange format .

Answer 6

1/(1+x^3)=1/(1+x)(1-x+x^2) resolve into partial fractions
let 1/(1+x)(1-x+x^2) be=A/(1+x)+(Bx+C)/(1-x+x^2)
A(1-x+x^2)+(Bx+C)(1+x)=1
A-Ax+Ax^2+Bx+Bx^2+C+Cx=1
A+C=1
-A+B+C=0
A+B=0
solving B+2C=1 and
2B+C=1
-2B-4C=-2
-3C=-1 C=1/3 A=2/3 and B=-2/3
so 1+x^3=2/3(1+x)+ (-2/3x+1/3)/(1-x+x^2)
=>[(2/3)(1/(1+x)]-[(2/3x)(1/(1-x+x^2)]+[(1/3)/1/(1-x+x^2)]
all are standard integrals.integrate and apply the limits

Answer 7

a million. First trick is to parametrize in terms of t. so in the event that they supply you a line necessary of ds, you may desire to restate it as dt. in this, you may know ds=?(?x/?t)^2+(?y/?t)^2 2. If the required is over a closed curve C, then you somewhat can use eco-friendly's Theorem. 3. in the event that they supply you the function as ?Pdx+Qdy, you could rewrite it in terms of ?F?dr the place F = F(r(t)) and dr = r'(t)dt. of course you get r(t) out of your curve C and F(x,y) = so F(r(t)) =

Answer 8

go to a good bookstore, probably a college bookstore
ask if they have Schaum's outline series
buy the one for Calculus..

a great reference work....
then, you can do your own [home]work

Answer 9

-81/4

Answer 10

it is super ugly, involving inverse tan, some radicals, pi, natural logs, and some fractions