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# Integral help please [Solved!]

### My question

can you help me solve this problem?

int0^1 1/(1+x^4)dx

### Relevant page

8. Integration by Trigonometric Substitution

### What I've done so far

I've looked all over the techniques in the Methods of Integration chapter, but cant find anything that will do it.

X

can you help me solve this problem?

int0^1 1/(1+x^4)dx
Relevant page

<a href="/methods-integration/8-integration-trigonometric-substitution.php">8. Integration by Trigonometric Substitution</a>

What I've done so far

I've looked all over the techniques in the Methods of Integration chapter, but cant find anything that will do it.

## Re: Integral help please

Hi Garrett

It is possible, but it is pretty ugly.

You need to express 1+x^4 as

(x^2+x sqrt2 + 1)(x^2-x sqrt2 +1)

Then separate it into 2 fractions using partial fractions techniques:

11. Integration by Partial Fractions

Your final answer involves the log of the above expressions in brackets and arctan of a conjugate pair.

Personally, I would suggest approaching it using numerical integration:

6. Simpson’s Rule

It is less work, less agony and your final answer will have similar decimal place accuracy. But I guess you are required to use algebraic integration techniques, yes?

BTW, your final answer is around 0.87.

X

Hi Garrett

It is possible, but it is pretty ugly.

You need to express 1+x^4 as

(x^2+x sqrt2 + 1)(x^2-x sqrt2 +1)

Then separate it into 2 fractions using partial fractions techniques:

<a href="/methods-integration/11-integration-partial-fractions.php">11. Integration by Partial Fractions</a>

Your final answer involves the log of the above expressions in brackets and arctan of a conjugate pair.

Personally, I would suggest approaching it using numerical integration:

<a href="/integration/6-simpsons-rule.php">6. Simpson&rsquo;s Rule</a>

It is less work, less agony and your final answer will have similar decimal place accuracy. But I guess you are required to use algebraic integration techniques, yes?

BTW, your final answer is around 0.87.

## Re: Integral help please

Thanks for your help sir, but I don't understand how you express 1+x^4 as (x^2+x sqrt2 + 1)(x^2-x sqrt2 +1). Can you show me how to do that ?

(and yes, i have to do it algebraically)

X

Thanks for your help sir, but I don't understand how you express 1+x^4 as (x^2+x sqrt2 + 1)(x^2-x sqrt2 +1). Can you show me how to do that ?

(and yes, i have to do it algebraically)

## Re: Integral help please

Hi Garrett

To get an x^4+1 term, it will involve the expansion of (x^2 + 1)^2.

But when you do that, you don't quite get (x^4+1) because you will have 2x^2 that you don't want. You can remove that term by including the +xsqrt2 and -xsqrt2 terms as shown.

It might be easier to think of it as:

[(x^2 + 1)+x sqrt2] [(x^2 +1) - x sqrt2]

We are just using difference of 2 squares:

2. Common Factor and Difference of Squares

Hope that helps.

X

Hi Garrett

To get an x^4+1 term, it will involve the expansion of (x^2 + 1)^2.

But when you do that, you don't quite get (x^4+1) because you will have 2x^2 that you don't want. You can remove that term by including the +xsqrt2 and -xsqrt2 terms as shown.

It might be easier to think of it as:

[(x^2 + 1)+x sqrt2] [(x^2 +1) - x sqrt2]

We are just using difference of 2 squares:

<a href="/factoring-fractions/2-common-factor-difference-squares.php">2. Common Factor and Difference of Squares</a>

Hope that helps.

## Re: Integral help please

ok, I got that part now.

For the integration, you said to do partial fractions.

1/((x^2+x sqrt2 + 1)(x^2-x sqrt2 +1))

=A/(x^2+x sqrt2 + 1)+ B/ (x^2 - x sqrt2 + 1)

So

x^4+1  = A(x^2 - x sqrt2 + 1) +B(x^2 + x sqrt2 + 1)

But I'm stuck there.

X

ok, I got that part now.

For the integration, you said to do partial fractions.

1/((x^2+x sqrt2 + 1)(x^2-x sqrt2 +1))

=A/(x^2+x sqrt2 + 1)+ B/ (x^2 - x sqrt2 + 1)

So

x^4+1  = A(x^2 - x sqrt2 + 1) +B(x^2 + x sqrt2 + 1)

But I'm stuck there.

## Re: Integral help please

There are a few problems. Once we do the cross multiply, we'll just have 1 on the LHS.

Hint 1: Our aim is to eventually get the number 1 on the RHS as well. (This is the 1 on the top of 1/(x^4+1).

Hint 2: When you have the highest power of x in the denominator being 2, what will you put in the top of your 2 partial fractions?

Can you now set up an expression and then solve it for A and B?

X

There are a few problems. Once we do the cross multiply, we'll just have 1 on the LHS.

<b>Hint 1:</b> Our aim is to eventually get the number 1 on the RHS as well. (This is the 1 on the top of 1/(x^4+1).

<b>Hint 2:</b> When you have the highest power of x in the denominator being 2, what will you put in the top of your 2 partial fractions?

Can you now set up an expression and then solve it for A and B?

## Re: Integral help please

Ah, I see. So I guess it's

1/(x^4+1)=(Ax+B)/(x^2+x sqrt2 + 1) + (Cx+D)/ (x^2 - x sqrt2 + 1)

Multiplying out:

1  = (Ax+B)(x^2 -x sqrt2 + 1) + (Cx+D) (x^2 + x sqrt2 + 1)

1  = (Ax^3 -Ax^2 sqrt2 + Ax + Bx^2 -Bx sqrt2 + B) + (Cx^3 + Cx^2 sqrt2 + Cx + Dx^2 + Dx sqrt2 + D)

1  = (Ax^3 +(B-A sqrt2)x^2 + (A -B sqrt2)x + B) + (Cx^3 + (C sqrt2 +D)x^2 + (C + D sqrt2)x + D)

Then for the x^3 terms:

A + C =0 .... (1)

For the x^2 terms:

B - A sqrt2 + C sqrt2 + D = 0 .... (2)

For the x terms:

A -B sqrt2 + C + D sqrt2 = 0 .... (3)

For the constant terms:

B + D = 1 .... (4)

From (1), (2) and (4) I gott:

A = 1/(2sqrt(2)

C = -1/(2sqrt(2)

From (2), (3) and (4), I got

B=1/2 and D=1/2

So

1/(x^4+1)  =(xsqrt(2) + 2)/(4(x^2+1+xsqrt(2))) + (-xsqrt(2) + 2)/(4(x^2+1-xsqrt(2)))

Now the integral:

int 1/(x^4+1)dx  =int(xsqrt(2) + 2)/(4(x^2+1+xsqrt(2)))dx + int(-xsqrt(2) + 2)/(4(x^2+1-xsqrt(2)))dx

But I'm stuck again. What to do now?

X

Ah, I see. So I guess it's

1/(x^4+1)=(Ax+B)/(x^2+x sqrt2 + 1) + (Cx+D)/ (x^2 - x sqrt2 + 1)

Multiplying out:

1  = (Ax+B)(x^2 -x sqrt2 + 1) + (Cx+D) (x^2 + x sqrt2 + 1)

1  = (Ax^3 -Ax^2 sqrt2 + Ax + Bx^2 -Bx sqrt2 + B) + (Cx^3 + Cx^2 sqrt2 + Cx + Dx^2 + Dx sqrt2 + D)

1  = (Ax^3 +(B-A sqrt2)x^2 + (A -B sqrt2)x + B) + (Cx^3 + (C sqrt2 +D)x^2 + (C + D sqrt2)x + D)

Then for the x^3 terms:

A + C =0 .... (1)

For the x^2 terms:

B - A sqrt2 + C sqrt2 + D = 0 .... (2)

For the x terms:

A -B sqrt2 + C + D sqrt2 = 0 .... (3)

For the constant terms:

B + D = 1 .... (4)

From (1), (2) and (4) I gott:

A = 1/(2sqrt(2)

C = -1/(2sqrt(2)

From (2), (3) and (4), I got

B=1/2 and D=1/2

So

1/(x^4+1)  =(xsqrt(2) + 2)/(4(x^2+1+xsqrt(2))) + (-xsqrt(2) + 2)/(4(x^2+1-xsqrt(2)))

Now the integral:

int 1/(x^4+1)dx  =int(xsqrt(2) + 2)/(4(x^2+1+xsqrt(2)))dx + int(-xsqrt(2) + 2)/(4(x^2+1-xsqrt(2)))dx

But I'm stuck again. What to do now?

## Re: Integral help please

It's very good what you've done.

In your final steps to tidy up our partial fractions, you did a bit too much work!

I expect you got this as one of your intermediate steps:

1/(x^4+1)  =(x + sqrt(2))/(4sqrt(2)(x^2+1+xsqrt(2))) + (-x + sqrt(2))/(4sqrt(2)(x^2+1-xsqrt(2)))

Can you see it now?

X

It's very good what you've done.

In your final steps to tidy up our partial fractions, you did a bit too much work!

I expect you got this as one of your intermediate steps:

1/(x^4+1)  =(x + sqrt(2))/(4sqrt(2)(x^2+1+xsqrt(2))) + (-x + sqrt(2))/(4sqrt(2)(x^2+1-xsqrt(2)))

Can you see it now?

## Re: Integral help please

Of course!

int 1/(x^4+1)dx  =1/(4sqrt(2)) int(x + sqrt(2))/(x^2+1+xsqrt(2))dx + 1/(4sqrt(2)) int(-x + sqrt(2))/((x^2+1-xsqrt(2)))dx

=1/(4sqrt(2))[ ln(x^2+1+xsqrt(2)):}  {:- ln(x^2+1-xsqrt(2))]

=1/(4sqrt(2))[ ln(x^2+1+xsqrt(2)) / ln(x^2+1-xsqrt(2))]

For the definite integral we have:

int0^1 1/(x^4+1)dx =1/(4sqrt(2))[ ln(x^2+1+xsqrt(2)) / ln(x^2+1-xsqrt(2))]0^1

=1/(4sqrt(2))[ ln(2 + sqrt(2)) / ln(2 - sqrt(2)) - 1]

=0.86697

Thanks a lot for your help, sir!

X

Of course!

int 1/(x^4+1)dx  =1/(4sqrt(2)) int(x + sqrt(2))/(x^2+1+xsqrt(2))dx + 1/(4sqrt(2)) int(-x + sqrt(2))/((x^2+1-xsqrt(2)))dx

=1/(4sqrt(2))[ ln(x^2+1+xsqrt(2)):}  {:- ln(x^2+1-xsqrt(2))]

=1/(4sqrt(2))[ ln(x^2+1+xsqrt(2)) / ln(x^2+1-xsqrt(2))]

For the definite integral we have:

int0^1 1/(x^4+1)dx =1/(4sqrt(2))[ ln(x^2+1+xsqrt(2)) / ln(x^2+1-xsqrt(2))]0^1

=1/(4sqrt(2))[ ln(2 + sqrt(2)) / ln(2 - sqrt(2)) - 1]

=0.86697

Thanks a lot for your help, sir!

## Re: Integral help please

You're welcome. Well done!

X

You're welcome. Well done!

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