Question

Assertion :$\int \frac{x^4+1}{x^6+1}dx={\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}{x}^3+C$ Reason: $\int \frac{\cos x+x\sin x}{x(x+\cos x)}dx=\log (x^2+x\cos x)+C$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

Assertion(A): $\int \frac{x^4+1}{x^6+1}dx$
$=\int \frac{(x^4+1-x^2)+x^2}{(x^4-x^2+1)(x^2+1)}dx$
$=\int \frac{dx}{1+x^2}+\int \frac{x^{2}dx}{1+{\left(x^3\right)}^2}$
Put $x^3=t$ in the second integral
$\Rightarrow x^{2}dx=\frac{dt}{3}$
$={\tan }^{-1}x+\frac{1}{3}\int \frac{dt}{1+t^2}$
$={\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}{x}^3+C$
$\therefore$ Assertion (A) is correct.
Reason(R): $\int \frac{\cos x+x\sin x}{x(x+\cos x)}dx$
$=\int \frac{(\cos x+x)-x(1-\sin x)}{x(x+\cos x)}dx$
$=\int \frac{1}{x}dx-\int \frac{1-\sin x}{(x+\cos x)}dx$
Put $x+\cos x=t$
$\Rightarrow (1-\sin x)dx=dt$
$=\log x-\int \frac{dt}{t}$
$=\log x-\log (x+\cos x)+C$
$=\log \left(\frac{x}{x+\cos x}\right)+C$
$\therefore$ Reason (R) is false.