Solving Integrals with Substitution Method - Step-by-Step Guide

[Nyasha]

Homework Statement

In each of the following cases use the given substitution in order to evaluate the given integral:

\int\frac{2+\sqrt x}{1-\sqrt x}dx=2(4(1+\sqrt x)^\frac{3}{2}-\frac{(1-\sqrt x)^2}{2}-3In(1-\sqrt x)

Homework Equations

For the substitution
u=1-\sqrt x

x=(1-u)^2

The Attempt at a Solution

du=\frac{-1}{2\sqrt x}

-2\sqrt x du=dx

-2\sqrt(1-u)^2du=dx


-2\int\frac{2+\sqrt(1-u)^2}{u}(\sqrt(1-u)^2

-2\int 2\sqrt(1-u^2)(u^-1)du+ \sqrt(1-u^2)(u^-1)

Guys how come when l simplify this integration l do not get the answer correct answer which is shown in the problem statement ? Where have l done my mistake ?

 

[gabbagabbahey]

du=\frac{-1}{2\sqrt x}

Surely you mean \frac{du}{dx}=\frac{-1}{2\sqrt x}...right?

-2\sqrt x du=dx

-2\sqrt(1-u)^2du=dx

Technically, this is incorrect; since sqrt() always reteurns the positive root,

\sqrt{(1-u)^2}=\left\{\begin{array}{lr}1-u, & 1-u \geq 0 \\ u-1, & 1-u\leq 0\end{array}

But it should be clear from your definition of u that u=1-\sqrt{x}\implies \sqrt{x}=1-u always. So \sqrt{x}\neq\sqrt{(1-u)^2} in general.

-2\int 2\sqrt(1-u^2)(u^-1)du+ \sqrt(1-u^2)(u^-1)

Now you seem to be claiming \sqrt{(1-u)^2}=\sqrt{1-u^2}...surely you didn't mean to do that!

Use \sqrt{x}=1-u instead.

 

[Nyasha]

Surely you mean \frac{du}{dx}=\frac{-1}{2\sqrt x}...right?



Technically, this is incorrect; since sqrt() always reteurns the positive root,

\sqrt{(1-u)^2}=\left\{\begin{array}{lr}1-u, & 1-u \geq 0 \\ u-1, & 1-u\leq 0\end{array}

But it should be clear from your definition of u that u=1-\sqrt{x}\implies \sqrt{x}=1-u always. So \sqrt{x}\neq\sqrt{(1-u)^2} in general.




Now you seem to be claiming \sqrt{(1-u)^2}=\sqrt{1-u^2}...surely you didn't mean to do that!

Use \sqrt{x}=1-u instead.

Maybe l should take a new approach since this one was a complete disaster.

x=(1-u)^2

dx= -2(1-u)du

dx=2u-2du

\int\frac{2+\sqrt (1-u)^2}{u}.(2u-2)du

I would like to know if l am on the right path before l take another wild goose chase

 

[gabbagabbahey]

Maybe l should take a new approach since this one was a complete disaster.

x=(1-u)^2

dx= -2(1-u)du

dx=2u-2du

\int\frac{2+\sqrt (1-u)^2}{u}.(2u-2)du

I would like to know if l am on the right path before l take another wild goose chase

Again, \sqrt{(1-u)^2} depends on whether 1-u is positive or negative.

To avoid having to analyze the two different cases, stick to your original substitution; u=1-\sqrt{x}\implies \sqrt{x}=1-u...

 

[Nyasha]

Again, \sqrt{(1-u)^2} depends on whether 1-u is positive or negative.

To avoid having to analyze the two different cases, stick to your original substitution; u=1-\sqrt{x}\implies \sqrt{x}=1-u...

Ohhh, l now get it. This means my integrand will become:


<br /> \int\frac{2+ (1-u)}{u}(2u-2)du=\int\frac{4u-2}{u}du + \int\frac{(2u-2)(1-u)}{u}du<br />

 

[gabbagabbahey]

Ohhh, l now get it. This means my integrand will become:


<br /> \int\frac{2+ (1-u)}{u}(2u-2)du=\int\frac{4u-2}{u}du + \int\frac{(2u-2)(1-u)}{u}du<br />

Actually, \int\frac{2+ (1-u)}{u}(2u-2)du=\int\frac{4u-4}{u}du + \int\frac{(2u-2)(1-u)}{u}du

But there is a much easier way to simplify this:

\int\frac{2+ (1-u)}{u}(2u-2)du=2\int\frac{(3-u)(u-1)}{u}du

And (3-u)(u-1)=___?

 

[Nyasha]

Actually, \int\frac{2+ (1-u)}{u}(2u-2)du=\int\frac{4u-4}{u}du + \int\frac{(2u-2)(1-u)}{u}du

But there is a much easier way to simplify this:

\int\frac{2+ (1-u)}{u}(2u-2)du=2\int\frac{(3-u)(u-1)}{u}du

And (3-u)(u-1)=___?

Thanks very much for the help