Worked solutions seem wrong.... (2 Viewers)

Run hard@thehsc · Jun 5, 2022

Thanks (for part i)

5uckerberg · Jun 5, 2022

Did you integrate 1 to get x and differentiated the rest?

I find that students will commonly make this mistake when they do integration by parts. Over complicate the integral
Next, bang their heads on the wall
Then cry for help because they had a hard time with the question.

ExtremelyBoredUser · Jun 5, 2022

Run hard@thehsc said:
View attachment 35723
Thanks (for part i)

Let $u = (1+x^2)^{-n}$ so $du = -2nx(1+x^2)^{-n-1} \, dx$ and $dv = dx$ so $v = x$ :

$I_{n} = \left[\frac{x}{(x^2+1)^{n}}\right]_{0}^{1} + 2n \int_{0}^{1} x^{2}(1+x^2)^{-n-1} \, dx$
$I_{n} = 2^{-n} + 2n \int_{0}^{1} (x^{2}+1)(1+x^2)^{-n-1} - (1+x^{2})^{-n-1} \, dx$
$I_{n} = 2^{-n} + 2n\left[I_{n} - I_{n+1}\right]$
$-2^{-n} + I_{n} -2nI_{n} = - 2n(I_{n+1})$
$I_{n+1} = n2^{-n-1} + \frac{2n-1}{2n} I_{n}$

5uckerberg · Jun 5, 2022

5uckerberg said:
Did you integrate 1 to get x and differentiated the rest?

I find that students will commonly make this mistake when they do integration by parts. Over complicate the integral
Next, bang their heads on the wall
Then cry for help because they had a hard time with the question.

Instead as @ExtremelyBoredUser has shown one should look at the how the equation is formatted and then work backwards.

Worked solutions seem wrong.... (2 Viewers)

Run hard@thehsc

Well-Known Member

5uckerberg

Well-Known Member

ExtremelyBoredUser

Bored Uni Student

5uckerberg

Well-Known Member

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