Question

If \[ \int (1+x)\log(1+x^2)\,dx= \left(x+\frac{x^2}{2}+\frac{1}{2}\right)\log(1+x^2)+g(x)+C, \] then \(g(x)=\)

• A. \(-2x-\dfrac{x^2}{2}+2\tan^{-1}x\)
• B. \(2\tan^{-1}x+\dfrac{x^2}{2}+\dfrac{x^3}{3}\)
• C. \(2\tan^{-1}x-\dfrac{x^2}{2}+3x\)
• D. \(2\tan^{-1}x+3x+\dfrac{x^3}{2}\)

Hint:

When an integral is already given in a special form, differentiate the given result and compare both sides to find the unknown function.

Answer 1

The Correct Option is A

Step 1: Compare the given integral form.
Given, \[ \int (1+x)\log(1+x^2)\,dx= \left(x+\frac{x^2}{2}+\frac{1}{2}\right)\log(1+x^2)+g(x)+C \] Differentiate both sides with respect to \(x\).
\[ (1+x)\log(1+x^2) = (1+x)\log(1+x^2) + \left(x+\frac{x^2}{2}+\frac{1}{2}\right)\frac{2x}{1+x^2} + g'(x) \]

Step 2: Find \(g'(x)\).
Canceling \((1+x)\log(1+x^2)\) from both sides, we get \[ g'(x)=- \left(x+\frac{x^2}{2}+\frac{1}{2}\right)\frac{2x}{1+x^2} \] \[ g'(x)=- \frac{x(x^2+2x+1)}{1+x^2} \] \[ g'(x)=- \frac{x^3+2x^2+x}{1+x^2} \]

Step 3: Simplify \(g'(x)\).
Now, \[ \frac{x^3+2x^2+x}{1+x^2} = x+2-\frac{2}{1+x^2} \] Therefore, \[ g'(x)=- \left(x+2-\frac{2}{1+x^2}\right) \] \[ g'(x)=-x-2+\frac{2}{1+x^2} \]

Step 4: Integrate to find \(g(x)\).
\[ g(x)=\int\left(-x-2+\frac{2}{1+x^2}\right)\,dx \] \[ g(x)=-\frac{x^2}{2}-2x+2\tan^{-1}x \] Thus, \[ g(x)=-2x-\frac{x^2}{2}+2\tan^{-1}x \]

Step 5: Final conclusion.
Therefore, \[ \boxed{-2x-\frac{x^2}{2}+2\tan^{-1}x} \]