help me to solve a calculus problem

[nn2000]

evaluate the intergral for dx/(1+x*x*x*x)

 

[5489]

Originally posted by: nn2000
evaluate the intergral for dx/(1+x*x*x*x)

is that x to the 4th power?

 

[RabidMongoose]

The answer is apple pie.

 

[Chu]

Originally posted by: nn2000
evaluate the intergral for dx/(1+x*x*x*x)

use trig substitution

EDIT : n/m

 

[Evadman]

Text

 

[nn2000]

Originally posted by: 5489

Originally posted by: nn2000
evaluate the intergral for dx/(1+x*x*x*x)

is that x to the 4th power?

yes, x to the 4th power

 

[gentobu]

Is it:


integral (1/(1+(x^2)^2)dx = tan^(-1)x^2+C ?

 

[Chu]

Originally posted by: nn2000
evaluate the intergral for dx/(1+x*x*x*x)

Wh00t, haven't had calc in 3 years even though I'm a MA major, but think I got it.

(1/(x^4+1)) = (1/(x^2+1)) * (x^2+1)/(x^4+1)

Then use the muplicand rule, keeping in mind int(1/(x^2+1) dx) = arctan(x)

EDIT : don't really need to use the arctan(x) equation, I started it out on paper. It's MESSY, but doable. Forgot what the muplicand rule is really called, it's the uv-int(v*du) rule.

 

[apac]

42.

for future reference, x to the 4th power can be written 'x^4'

 

[nn2000]

Originally posted by: Chu

Originally posted by: nn2000
evaluate the intergral for dx/(1+x*x*x*x)

Wh00t, haven't had calc in 3 years even though I'm a MA major, but think I got it.

(1/(x^4+1)) = (1/(x^2+1)) * (x^2+1)/(x^4+1)

Then use the muplicand rule, keeping in mind int(1/(x^2+1) dx) = arctan(x)

EDIT : don't really need to use the arctan(x) equation, I started it out on paper. It's MESSY, but doable. Forgot what the muplicand rule is really called, it's the uv-int(v*du) rule.

so u = (1/(x^2+1)), dv = (x^2+1)/(x^4+1) or vice versa? both are messy, but let me try it.

 

[YankeezXBA]

after this we're going to help you to solve your english problem

 

[cain]

it's called mathematica