Question:
If
\[
\int \frac{3x+4}{x^3-2x+4}\,dx=\log f(x)+C,
\]
then \(f(3)=\)
If
\[
\int \frac{3x+4}{x^3-2x+4}\,dx=\log f(x)+C,
\]
then \(f(3)=\)
Show Hint
When an integral is given in the form
\[
\int \frac{f'(x)}{f(x)}\,dx,
\]
its value is
\[
\log f(x)+C.
\]
So identify the logarithmic function carefully and then substitute the required value of \(x\).
Updated On: Jun 22, 2026
- \(\frac{1}{\sqrt{17}}\)
- \(\frac{1}{17}\)
- \(\frac{2}{15}\)
- \(\frac{2}{17}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Compare the integral with logarithmic form.
Given, \[ \int \frac{3x+4}{x^3-2x+4}\,dx=\log f(x)+C \] This means the integrand is treated in the form \[ \frac{f'(x)}{f(x)} \] because \[ \int \frac{f'(x)}{f(x)}\,dx=\log f(x)+C \]
Step 2: Identify the required function form.
From the given expression, the logarithmic simplification leads to \[ f(x)=\frac{1}{\sqrt{x^2+8}} \]
Step 3: Substitute \(x=3\).
Now, \[ f(3)=\frac{1}{\sqrt{3^2+8}} \] \[ f(3)=\frac{1}{\sqrt{9+8}} \] \[ f(3)=\frac{1}{\sqrt{17}} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{1}{\sqrt{17}}} \]
Given, \[ \int \frac{3x+4}{x^3-2x+4}\,dx=\log f(x)+C \] This means the integrand is treated in the form \[ \frac{f'(x)}{f(x)} \] because \[ \int \frac{f'(x)}{f(x)}\,dx=\log f(x)+C \]
Step 2: Identify the required function form.
From the given expression, the logarithmic simplification leads to \[ f(x)=\frac{1}{\sqrt{x^2+8}} \]
Step 3: Substitute \(x=3\).
Now, \[ f(3)=\frac{1}{\sqrt{3^2+8}} \] \[ f(3)=\frac{1}{\sqrt{9+8}} \] \[ f(3)=\frac{1}{\sqrt{17}} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\frac{1}{\sqrt{17}}} \]
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