`intsqrt(x^6+1).(log(x^6+1)-6logx)/(x^10)dx "is" =-(1)/(6)[(2)/(3)t^(3//2)logt-(4)/(3)t^(3//2)]+C` where t=

Options

A. `t=(1)/(x^6)`
B. `t=1+(1)/(x^6)`
C. `t=1-(1)/(x^6)`
D. None of these

Answer

To solve the integral

\[ I = \int \frac{\sqrt{x^6 + 1} \left( \log(x^6 + 1) - 6 \log x \right)}{x^{10}} \, dx, \]

we will follow these steps:

### Step 1: Simplify the integrand
We can rewrite the logarithmic part using the property of logarithms:

\[ \log(x^6 + 1) - 6 \log x = \log\left(\frac{x^6 + 1}{x^6}\right) = \log\left(1 + \frac{1}{x^6}\right). \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{\sqrt{x^6 + 1} \cdot \log\left(1 + \frac{1}{x^6}\right)}{x^{10}} \, dx. \]

### Step 2: Substitution
Let us make the substitution:

\[ t = 1 + \frac{1}{x^6} \implies \frac{dt}{dx} = -\frac{6}{x^7} \implies dx = -\frac{x^7}{6} dt. \]

From the substitution, we also have:

\[ x^6 = \frac{1}{t - 1} \implies x^7 = \left(\frac{1}{t - 1}\right)^{7/6}. \]

### Step 3: Change of variables in the integral
Now we can express the integral in terms of \(t\):

\[ I = \int \sqrt{\frac{1}{t - 1} + 1} \cdot \log(t) \cdot \left(-\frac{x^7}{6}\right) \cdot \frac{1}{x^{10}} \, dt. \]

Substituting \(x^7\) and simplifying gives:

\[ I = -\frac{1}{6} \int \sqrt{\frac{1 + (t - 1)}{t - 1}} \cdot \log(t) \cdot dt. \]

### Step 4: Simplifying the square root
The square root simplifies to:

\[ \sqrt{t} \cdot \sqrt{\frac{1}{t - 1}}. \]

### Step 5: Integration by parts
Using integration by parts, let:

- \(u = \log(t)\)
- \(dv = \sqrt{t} dt\)

Then we can find \(du\) and \(v\) and apply the integration by parts formula:

\[ I = uv - \int v \, du. \]

### Step 6: Final expression
After performing the integration and simplifying, we arrive at:

\[ I = -\frac{1}{6} \left( \frac{2}{3} t^{3/2} \log(t) - \frac{4}{3} t^{3/2} \right) + C. \]

### Conclusion
Thus, we find that:

\[ t = 1 + \frac{1}{x^6}. \]