Integral with trig substitution

In summary, for the given integral, you can separate it into two parts: \displaystyle \int \frac{x}{(x^2+1)^2}\,dx and \displaystyle \int \frac{1}{(x^2+1)^2}\,dx. For the first part, use substitution to get the result \frac{1}{2}\arctan(x). For the second part, use integration by parts to get the result \frac{x-1}{2(x^2+1)} + C.
  • #1
hahaha158
80
0

Homework Statement



∫(x+1)/((x^2+1)^2)

Homework Equations


The Attempt at a Solution



I have been able to separate this into 2

∫x/(x^2+1)^2 dx which i found to be equal to (1/2)arctanx

and

∫1/(x^2+1)^2 dx which i am unable to find

What i did was sub in x=tanθ and dx=sec^2(θ)dθ, and with this i was able to get the integral to the form

∫1/sec^2(θ)dθ=∫cos^2(θ)dθ =

∫(1+cos2θ)/2

I then separated this into two integrals

∫1/2=θ/2

and

∫cos2θ/2 = sin2θ/4

However, the answer given is (x-1)/(2(x^2+1)+C

I am not sure how to get to this answer from

θ/2+sin2θ/4+C, i tried subbing in θ=arctanx and i get a very messy equation

arctanx/2+sin(2arctanx)/4+C

I was able to simplify it to

cosx/2sinx+sin(2cosx/sinx)/4+C, and i can't get any farther twoards the answer.

Can anyone please explain how i get there or did i make a mistake somewhere in my work?

Thanks very much
 
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  • #2
For what it's worth, their answer is certainly wrong because it surely must have an arctan(x) term. Maple gives:

[(1/4)(4x-2)/(x^2+1)] + arctan(x)
 
  • #3
LCKurtz said:
For what it's worth, their answer is certainly wrong because it surely must have an arctan(x) term. Maple gives:

[(1/4)(4x-2)/(x^2+1)] + arctan(x)

oops i entered the original question incorrectly, it should be 1+x on the top rather than 2+x, also the answer i wrote was for one part of the integral only, not the ansewr to the whole question, i should have clariified sorry.
 
  • #4
hahaha158 said:

Homework Statement



∫(x+1)/((x^2+1)^2)

Homework Equations



The Attempt at a Solution



I have been able to separate this into 2

∫x/(x^2+1)^2 dx which i found to be equal to (1/2)arctanx

and

∫1/(x^2+1)^2 dx which i am unable to find
...

Thanks very much
[itex]\displaystyle \int \frac{x}{(x^2+1)^2}\,dx\ne\frac{1}{2}\arctan(x)[/itex]

Use substitution to evaluate this integral. Let u = x2 + 1 ...


For [itex]\displaystyle \int \frac{1}{(x^2+1)^2}\,dx\,,\ [/itex] write the integrand as [itex]\displaystyle \frac{x^2+1}{(x^2+1)^2}-\frac{x^2}{(x^2+1)^2}\ .\ [/itex]

The first term gives the arctan part of the result. For the second term, do integration by parts, with u = x, dv = ...
 

1. What is "integral with trig substitution"?

Integral with trig substitution is a method used in calculus to evaluate integrals that involve trigonometric functions. It involves substituting a trigonometric expression for a variable in the integral in order to simplify the integration process.

2. When should I use "integral with trig substitution"?

Trig substitution is useful when you have an integral that involves a radical expression, especially when the radicand (expression under the radical) contains a sum or difference of squares. It can also be used when the integrand contains a quadratic expression that cannot be factored.

3. How do I perform "integral with trig substitution"?

To perform trig substitution, you first need to identify which trigonometric expression to substitute. This is typically determined by the form of the integrand. Once you have identified the substitution, you then use trigonometric identities to rewrite the integral in terms of the substituted variable, and then integrate as usual.

4. What are some common trig substitutions used?

Some common trig substitutions used include:

  • √(a²-x²) → x = a sinθ
  • √(a²+x²) → x = a tanθ
  • √(x²-a²) → x = a secθ
  • √(x²+a²) → x = a cotθ

5. What are the benefits of using "integral with trig substitution"?

Using trig substitution can often simplify the integral and make it easier to evaluate. It also allows us to solve integrals that would otherwise be difficult or impossible to solve using other methods. Additionally, it can be used to solve integrals involving trigonometric functions that cannot be solved using basic integration techniques.

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