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

∫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

\(\dfrac{1}{2}(∫logt dt)\)

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

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

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