`I=int(dx)/((a+dx^2)sqrt(b-a x^2))`

Options

A. `(1)/(sqrt(a(a^(2)+b^(2))))tan^(-1)((xsqrt(a^(2)+b^(2)))/(asqrt(b-ax^(2))))+C`
B. `(1)/(sqrt((a^(2)+b^(2))))tan^(-1)((xsqrt(a^(2)+b^(2)))/(asqrt(b-ax^(2))))+C`
C. `(1)/(sqrt(a(a^(2)+b^(2))))tan^(-1)((xsqrt(a^(2)+b^(2)))/a)+C`
D. None of the above

Answer

To solve the integral

\[ I = \int \frac{dx}{(a + dx^2) \sqrt{b - ax^2}}, \]

we will follow the steps outlined in the video transcript.

### Step 1: Substitution
Let \( x = \frac{1}{t} \). Then, we have:

\[ dx = -\frac{1}{t^2} dt. \]

### Step 2: Substitute in the Integral
Substituting \( x \) and \( dx \) into the integral gives:

\[ I = \int \frac{-\frac{1}{t^2} dt}{\left(a + d\left(\frac{1}{t}\right)^2\right) \sqrt{b - a\left(\frac{1}{t}\right)^2}}. \]

This simplifies to:

\[ I = -\int \frac{dt}{t^2 \left(a + \frac{d}{t^2}\right) \sqrt{b - \frac{a}{t^2}}}. \]

### Step 3: Simplify the Denominator
The denominator can be rewritten as:

\[ a + \frac{d}{t^2} = \frac{at^2 + d}{t^2}, \]

and

\[ \sqrt{b - \frac{a}{t^2}} = \sqrt{\frac{bt^2 - a}{t^2}} = \frac{\sqrt{bt^2 - a}}{t}. \]

Substituting these into the integral gives:

\[ I = -\int \frac{dt}{\frac{at^2 + d}{t^2} \cdot \frac{\sqrt{bt^2 - a}}{t}} = -\int \frac{t^3 dt}{(at^2 + d) \sqrt{bt^2 - a}}. \]

### Step 4: Further Substitution
Let \( z^2 = bt^2 - a \). Then, differentiating gives:

\[ 2z dz = b \cdot 2t dt \implies t dt = \frac{z dz}{b}. \]

### Step 5: Substitute Back into the Integral
Now, express \( t^2 \) in terms of \( z \):

\[ bt^2 = z^2 + a \implies t^2 = \frac{z^2 + a}{b}. \]

Substituting back into the integral gives:

\[ I = -\int \frac{z dz}{(a \cdot \frac{z^2 + a}{b} + d) \sqrt{z^2}}. \]

### Step 6: Integrate
Now we can simplify and integrate. The integral can be expressed in a standard form, and we can use the result:

\[ \int \frac{dz}{z^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{z}{a}\right). \]

### Final Result
After performing the integration and substituting back for \( z \) in terms of \( x \), we arrive at the final result:

\[ I = -\frac{1}{\sqrt{a}} \cdot \frac{1}{\sqrt{a + b}} \tan^{-1}\left(\frac{\sqrt{b - ax^2}}{\sqrt{a + b}}\right) + C, \]

where \( C \) is the constant of integration.