For integral `int f(x- a/x)*(1+a/x^(2))dx`, put `x-a/x=t`
For integral `int f(x+a/x)*(1-(a)/(x^(2)))dx`, put `x+a/x =t`
For integral `int f(x^(2)-(a)/(x^(2)))*(x+(a)/(x^(3)))dx`, put `x^(2)-(a)/(x^(2))=t`
For integral `int f(x^(2)+(a)/(x^(2)))*(x-(a)/(x^(3)))dx`, put `x^(2)+(a)/(x^(2))=t`
many integrands can be brought into above forms by suitable reductions or transformations .
`int(5x^(4)+4x^(5))/((x^(5)+x+1)^(2))dx`

Options

A. `x^(5)+x+1+C`
B. `(x^(5))/(x^(5)+x+1)+C`
C. `x^(-4)+x^(-5)+C`
D. `(x^(5))/(x^(5)+x+1)+C`

Answer

To solve the integral \( I = \int \frac{5x^4 + 4x^5}{(x^5 + x + 1)^2} \, dx \), we will follow these steps:

### Step 1: Simplify the Integral
First, we can factor out \( x^5 \) from the denominator:

\[ I = \int \frac{5x^4 + 4x^5}{(x^5 + x + 1)^2} \, dx = \int \frac{5x^4 + 4x^5}{x^{10}(1 + \frac{1}{x^4} + \frac{1}{x^5})^2} \, dx \]

### Step 2: Substitute for Simplicity
Let \( t = 1 + \frac{1}{x^4} + \frac{1}{x^5} \). Then, we need to find \( dt \):

\[ dt = -\left(\frac{4}{x^5} + \frac{5}{x^6}\right) dx \]

Rearranging gives:

\[ dx = -\frac{x^6}{4 + 5/x} dt \]

### Step 3: Rewrite the Integral
Now substituting \( t \) and \( dx \) into the integral:

\[ I = \int \frac{5x^4 + 4x^5}{(x^5 + x + 1)^2} \cdot -\frac{x^6}{4 + 5/x} dt \]

This simplifies to:

\[ I = -\int \frac{5x^4 + 4x^5}{t^2} \cdot \left(\frac{x^6}{4 + 5/x}\right) dt \]

### Step 4: Further Simplification
Now, we can express \( 5x^4 + 4x^5 \) in terms of \( t \):

\[ I = -\int \frac{(5 + 4x)x^4}{t^2} dt \]

### Step 5: Solve the Integral
The integral can now be expressed as:

\[ I = -\int \frac{1}{t^2} dt \]

The integral of \( t^{-2} \) is:

\[ I = \frac{1}{t} + C \]

### Step 6: Substitute Back for \( t \)
Substituting back for \( t \):

\[ I = \frac{1}{1 + \frac{1}{x^4} + \frac{1}{x^5}} + C = \frac{x^5}{x^5 + x + 1} + C \]

### Final Answer:
Thus, the integral evaluates to:

\[ I = \frac{x^5}{x^5 + x + 1} + C \]