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

Answer: (B)

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

Answer 2

∫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)