Question

\(∫xlog(1+x^2)dx=\)?

• A. \(\dfrac{1}{2}(1+x^2)log(2+x^2)+\dfrac{x^{2}}{2}+C\)
• B. \(\dfrac{1}{2}(1+x^2)log(1+x^2)-(\dfrac{1+x^2}{2})+C\)
• C. \(\dfrac{1}{2}(1+x^2)log(2+x^2)-\dfrac{x^{2}}{2}+C\)
• D. \((1+x^2)log(1+x^2)+(1+x^2)+C\)
• E. \((1-x^2)log(1+x^2)+(1-x^2)+C\)

Answer 1

The Correct Option is B

Step 1: Use integration by parts with: \[ u = \log(1+x^2) \quad \Rightarrow \quad du = \frac{2x}{1+x^2} dx \] \[ dv = x dx \quad \Rightarrow \quad v = \frac{x^2}{2} \]

Step 2: Apply the integration by parts formula: \[ \int u dv = uv - \int v du \] \[ \int x \log(1+x^2) dx = \frac{x^2}{2} \log(1+x^2) - \int \frac{x^2}{2} \cdot \frac{2x}{1+x^2} dx \]

Step 3: Simplify the remaining integral: \[ = \frac{x^2}{2} \log(1+x^2) - \int \frac{x^3}{1+x^2} dx \]

Step 4: Perform polynomial division on the integrand: \[ \frac{x^3}{1+x^2} = x - \frac{x}{1+x^2} \]

Step 5: Integrate term by term: \[ \int \left(x - \frac{x}{1+x^2}\right) dx = \frac{x^2}{2} - \frac{1}{2} \log(1+x^2) \]

Step 6: Combine all terms: \[ \frac{x^2}{2} \log(1+x^2) - \left(\frac{x^2}{2} - \frac{1}{2} \log(1+x^2)\right) + C \] \[ = \frac{1+x^2}{2} \log(1+x^2) - \frac{x^2}{2} + C \]

Conclusion: The correct answer is \(\boxed{B}\) \(\left( \frac{1}{2}(1+x^2)\log(1+x^2) - \frac{x^2}{2} + C \right)\).

Answer 2

Given

∫xlog(1+x^2)dx

To solve the question first multiply \(\dfrac{2}{2}\) in the above expression,

Then we get 

\(∫\dfrac{1}{2}×2xlog(1+x^2)dx\)

Now take ,\((1+x^2)= t\)

Now,  derivate both the sides with respect to \(x\) ,

Therefore, we get 

\(2x dx= dt\)

substituting this expression in the main (given) expression we get

\(\dfrac{1}{2}(∫logt dt)\)

\(=\dfrac{1}{2} (tlogt-t)+C\)

\(=\dfrac{1}{2}(1+x^{2})(log(1+x^{2})-1)\)

\(=\dfrac{1}{2}(1+x^{2})log(1+x^{2})-\dfrac{1+x^{2}}{2}\)