Sometimes a complicated looking integral can be reduced to a manageable one by just substituting a variable for other and reducing the integral. The new integral can be easily integrated and solution be found.  The integration by substitution sometimes called u-substitution because we normally used u for substituting is a method which is reverse (anti derivative) of chain rule of differentiation. The success depends on the choice of substitution.

$F(u)$ = $\int f(u)du$

$\frac{dF(u)}{du}$ = $f(u)$

Put $u$ = $\Psi(x)$ then $\frac{du}{dx}$ = $\Psi ' (x)$

$\frac{dF(u)}{dx}$ = $\frac{dF(u)}{du}$ $\times$ $\frac{du}{dx}$
 
$\frac{dF(u)}{dx}$ = $f(u) \times \Psi ' (x)$

$F(u)$ = $\int f(\Psi(x)) \Psi ' (x) dx$

$\int f(\Psi(x)) \Psi ' (x) dx$ = $\int f(u)du$

Some standard results of integration which use substitution method. Knowing these one can quickly identify which is going to be "$u$" .

$\int [f(x)]^n\ f' (x)dx$ = $\frac{[f(x)]^{(n+1)}}{(n+1)}$ $+\ C n \neq -1$

$\int$ $(\frac{(f ' (x))}{f(x)})$ $dx$ = $ln⁡(f(x)) + C$

$\int$ $(\frac{(f' (x))}{\sqrt{(f(x)}})$ $dx$ = $2 \sqrt x + C$ 

Let us prove these results using u-substitution 
1) $\int [f(x)]^n f' (x)dx$ = $\frac{[f(x)]^{(n + 1)}}{(n + 1)}$ $+\ Cn \neq -1$

Proof:  
 
$\int [f(x)]^n f' (x)dx$

Let $f(x)$ = $u$

$f' (x) dx$ = $du$

= $\int u^n du$

= $\frac{u^{(n + 1)}}{(n + 1)}$ $+\ C$

Substituting back for $u$

= $\frac{[f(x)]^{(n + 1)}}{(n + 1)}$ $+\ C$
2) $\int$ $(\frac{(f'(x)}{f(x)})$ $dx$ = $ln⁡(f(x)) + C$

Proof:  

$\int$ $(\frac{f' (x)}{f(x)})$ $dx$

Always select $f(x)$ as $u$

$f(x)$ = $u $

Differentiate $w.r.t.x$

$f' (x) dx$ = $du$

Now substitute in the given integral

$\int$ $\frac{du}{u}$
 
= $ln⁡(u) + C$

Substitute back for $u$

= $ln⁡(f(x)) + C$ 


3) $\int$ $(\frac{f' (x)}{\sqrt{(f(x)}})$ $dx$ = $2 \sqrt x + C$

Proof: 

Let $f(x)$ = $u$

$f' (x) dx$ = $du$

$\int$ $(\frac{du}{\sqrt u})$
 
= $\int$ $u^{\frac{-1}{2}}$ $du$

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

= $\frac{u^{\frac{1}{2}}}{\frac{1}{2}}$ $+\ C$

= $2u^{\frac{1}{2}}$ $+ C$

= $2 \sqrt u + C$

Substituting back for $u$ we get

= $2 \sqrt x + C$

Now time to look into some examples.

Example 1: 

Integrate $\int$ $\frac{2x + 1}{x^2 + x + 3}$ $dx$
 
Solution: 

First scan the problem to find which is $f(x)$ and if we have $f'(x)$ there as well.

So we can see the derivative of $x^2 + x + 3$  is $2x + 1$  Therefore, $x^2 + x + 3$  is our $u$.

Let $x^2 + x + 3$ = $u$

Differentiate with respect to $x$ we get

$2x + 1 dx$ = $du$

Our integration becomes

$\int$ $\frac{1}{u}$ $du$

= $ln⁡(u) + C$ 

Substituting back for $u$ we get

= $ln⁡(x^2 + x + 3) + C$
Example 2:

Integrate 

$\int$ $\frac{4x + 5}{\sqrt{2x^2 + 5x - 6}}$

Solution: 

In  this case check the derivative of the term under the square root and if it matches the other expression then we can select the term under the square root as $u$.

Let $2x^2 + 5x - 6$ = $u$

$(4x + 5)dx$ = $du$

The integral can be rewritten as 

$\int$ $\frac{du}{\sqrt u}$ 

= $\int$ $u^{-\frac{1}{2}}$ $du$

= $\frac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1}$  $+ C$

= $\frac{u^{\frac{1}{2}}}{\frac{1}{2}}$ $+ C$

= $2 \sqrt u + C$

Substituting back

= $2 \sqrt{(2x^2 + 5x - 6)} + C$
Example 3:  

$\int sin^6 (x) cos⁡(x) dx$

Solution:

Let $sin⁡(x)$ = $u$

Differentiate with respect to $x$.

$cos⁡(x)dx$ = $du$

The integral can be written as 

$\int u^6 du$

= $\frac{u^{(6 + 1)}}{(6 + 1)}$ $+ C$

= $\frac{u^7}{7}$ $+\ C$

Substituting back for $u$

= $\frac{1}{7}$ $sin^7\ (x) + C$ 

The above examples were from the results we had discussed now you would have the idea how those can be easily handled now let us take something different but using the same idea of substitution.
Example 4:  

Integrate $\int$ $\frac{(xsin^{-1} (x^2)}{\sqrt{1 - x^4}}$ $dx$

Solution:   

This might take a while and might require multiple substitution 

Let us start with u substitution

Let $x^2$ = $u$

Differentiate both sides with respect to $u$

$2x dx$ = $du$

$x dx$ = $\frac{du}{2}$ 

The integral can be written as 

$\int$ $\frac{sin^{-1} ⁡u}{\sqrt{ 1 - u^2}}$  $\frac{du}{2}$ 

= $\frac{1}{2}$ $\int$ $\frac{sin^{-1} u}{\sqrt{1 - u^2}}$ $du$

With experience one can easily now that this is of the form $\int$ $f(u)\ f' (u)du$

So let us use $v$ substitution as we have already used $u$.

Let $sin^{-1}\ (u)$ = $v$ 

Differentiate both sides with respect to $u$ we get

$\frac{1}{\sqrt{1 - u^2}}$ $du$ = $dv$

The integral now becomes

$\frac{1}{2}$ $\int\ v\ dv$

= $\frac{1}{4}$ $v^2 + C$ 

Substitute back for $v$

= $\frac{1}{4}$ $[sin^{-1} ⁡(u)]^2 + C$ 

Substitute back for $u$

= $\frac{1}{4}$ $[sin^{-1} ⁡(x^2 )]^2 + C$ 
Example 5:  

Integrate $\int$ $\frac{sin⁡(ln⁡(x))}{x}$ $dx$

Solution:

Let $ln⁡(x)$ = $u$

Differentiating with respect to $x$

$\frac{1}{x}$ $dx$ = $du$

Our integral becomes 

$\int$ $sin⁡(u)du$

= $-cos⁡(u) + C$ 

Substituting back for $u$

= $-cos⁡(ln⁡(x)) + C$