### 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.

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?

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.

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)

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.

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.

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?

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?

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?

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!

You're welcome. Well done!

X

You're welcome. Well done!

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