Depending on the exam board, after completing the square, you may be able to use some standard formulae given in the formula book which makes this question fairly easy. If not then the question is much longer.

Reply 4

Barney63

OP

Right I have done this question, but get a different answer to Wolfram Alpha.
I am the same as WA up until

\frac {1}{4} \int \frac {1} {\frac {u^2}{4} +1} du

What I have done is let

s=\frac{u}{4}

that giving me

Unparseable latex formula:

\frac{1}{4} \int \frac {1}{s^2 +1} du = \frac {1}{4} tan^-^1 \left( s \right) +C

subbing s and u back in

Unparseable latex formula:

\frac {1}{4} tan^-^1 \left( \frac {x+1}{4} \right) +C

WA on the other hand does let

s= \frac {u}{2}

and

ds = \frac {1}{2}

giving

Unparseable latex formula:

\frac {1}{2} tan^-^1 \left( s \right) +C

subbing s and u back in

Unparseable latex formula:

\frac {1}{2} tan^-^1 \left( \frac {x+1}{2} \right) +C

Why?

Barney

Reply 5

davros

16

Original post by Barney63

Right I have done this question, but get a different answer to Wolfram Alpha.
I am the same as WA up until

\frac {1}{4} \int \frac {1} {\frac {u^2}{4} +1} du

What I have done is let

s=\frac{u}{4}

that giving me

Unparseable latex formula:

\frac{1}{4} \int \frac {1}{s^2 +1} du = \frac {1}{4} tan^-^1 \left( s \right) +C

subbing s and u back in

Unparseable latex formula:

\frac {1}{4} tan^-^1 \left( \frac {x+1}{4} \right) +C

WA on the other hand does let

s= \frac {u}{2}

and

ds = \frac {1}{2}

giving

Unparseable latex formula:

\frac {1}{2} tan^-^1 \left( s \right) +C

subbing s and u back in

Unparseable latex formula:

\frac {1}{2} tan^-^1 \left( \frac {x+1}{2} \right) +C

Why?

Barney

why have you let s=u/4? that doesn't help you because then s^2 = u^2/16!
And you can't integrate a function of s with respect to u - you have to convert the whole integral consistently.

Reply 6

Barney63

OP

Sorry I didn't follow that.

Barney

Original post by Barney63

Sorry I didn't follow that.

Barney

You are making this much too hard. Why have you taken the factor of

\frac{1}{4}

out? This has created an unnecessary fractional denominator.

(edited 11 years ago)

Reply 8

Barney63

OP

\int \frac {1}{(x+1)^2+4}dx

then let u=x+1

\int \frac {1}{u^2+4}du

then

\int \frac{1}{4 \left( \frac {u^2}{4} +1 \right)} du

then

\frac {1}{4} \int \frac {1}{\frac{u^2}{4}+1} du

Or how would you recommend doing it?

Barney

(edited 11 years ago)

No substitution is necessary but, if you feel the need to make one, stop here and look at the standard integrals in your formula book.

Original post by Barney63

\displaystyle \int \frac {1}{(x+1)^2+4}\, dx

then let

u=x+1

\displaystyle \int \frac {1}{u^2+4}\, du

(edited 11 years ago)

Reply 10

Barney63

OP

We haven't been given/use a formula book :frown:

Barney

Reply 11

ztibor

10

Original post by Barney63

Right I have done this question, but get a different answer to Wolfram Alpha.
I am the same as WA up until

\frac {1}{4} \int \frac {1} {\frac {u^2}{4} +1} du

What I have done is let

s=\frac{u}{4}

You should to substitute

s=\frac{u}{2}

that giving me

Unparseable latex formula:

\frac{1}{4} \int \frac {1}{s^2 +1} du = \frac {1}{4} tan^-^1 \left( s \right) +C

1. Your substitution here is wrong totally
AS you wrote let s be

s=\frac{u}{4}

. OK.
then

u=4s ->u^2=16s^2 ->\frac{u^2}{4}=4s^2->(2s)^2

and you have to substitute the du factor too.
Without this, the integral in that form you wrote gives different funtion. Namely

\frac{1}{4} \int \frac {1}{s^2 +1} du =\frac {1}{s^2 +1}\cdot u+C

because s is a constant with respect to u. (integration factor is du here)
THe right method would be:
u=4s ->ratio of differentials

\frac{du}{ds}=4

->du=4ds

\frac{1}{4} \int \frac {1}{(2s)^2 +1} 4ds =\int \frac {1}{(2s)^2 +1}ds

and now use the t=2s substitution
from this

s=\frac{t}{2}->ds=\frac{1}{2}dt

and

\frac{1}{2} \int \frac{1}{1+t^2} dt

which is a base integral for t.
2. With

s=\frac{u}{2}

you will get similar form more simple way
then

u=2s->du=2ds

\frac{1}{4}\int \frac{1}{1+s^2} \cdot 2\cdot ds=\frac{1}{2}\int \frac{1}{1+s^2} ds

3. Most simple method without substitution is to use the known rules for
differentiation, composite function and the chain rule.

\displaystyle \frac{1}{4} \int \frac{1}{\left (\frac{x+1}{2}\right )^2+1}dx=

\displaystyle =\frac{1}{4} \cdot \frac{arc tan \left (\frac{x+1}{2}\right )}{\frac{1}{2}}= \frac{1}{2} \cdot arc tan \left (\frac{x+1}{2}\right )

(edited 11 years ago)

Original post by Barney63

We haven't been given/use a formula book :frown:

Barney

What course are you studying?

Reply 13

Barney63

OP

Its the Institute of Maths Polymaths course, an access course for the maths degree.

Barney

Original post by Barney63

Its the Institute of Maths Polymaths course, an access course for the maths degree.

Barney

Ok now I understand why you appeared to be making a meal of this.

A Level students are provided with a booklet containing this standard integral (along with many others):

\displaystyle \int \frac{1}{x^2+a^2} \, dx = \frac{1}{a}\tan^{-1}\frac{x}{a}+k

Reply 15

Barney63

OP

So that would give

\int \frac {1}{u^2+2^2}

Unparseable latex formula:

\frac{1}{2}tan^-^1 \frac {u}{2} +C

Barney

Original post by Barney63

So that would give

\int \frac {1}{u^2+2^2}

Unparseable latex formula:

\frac{1}{2}tan^-^1 \frac {u}{2} +C

Barney

Yes (if you remember the du at the end of your integral).

You might like to think about how you could prove this standard result (it isn't as difficult as you might think as you know the answer you are seeking).

Reply 17

yaboy

Original post by Mr M

Ok now I understand why you appeared to be making a meal of this.

A Level students are provided with a booklet containing this standard integral (along with many others):