The integral \int \frac(x^8-x^2) d x(x^12+3 x^6+1) \tan ^-1(x^3+\frac1x^3) is equal to :

\log _e(|\tan ^-1(x^3+\frac1x^3)|)^1 / 3+C

\log _e(|\tan ^-1(x^3+\frac1x^3)|)+C

\log _e(|\tan ^-1(x^3+\frac1x^3)|)^1 / 2+C

\log _e(|\tan ^-1(x^3+\frac1x^3)|)^3+C

Indefinite Integrals PYQ — JEE Main 2024 · 27 Jan · Shift 2 · Q47 with Solution

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Correct answer: A

Let $u=\tan^{-1}\left(x^3+\frac{1}{x^3}\right).$ Then the integral becomes of the form $\int \frac{u'}{u}\,dx,$ so we first compute $u'$ .
Differentiate the inner function: $\frac{d}{dx}\left(x^3+\frac{1}{x^3}\right)=3x^2-\frac{3}{x^4}=3\left(x^2-\frac{1}{x^4}\right)=\frac{3(x^6-1)}{x^4}.$

Hence, $u' = \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2}\cdot \frac{3(x^6-1)}{x^4}.$

= x^6+3+\frac{1}{x^6}.$$ So, $$u' = \frac{3(x^6-1)}{x^4\left(x^6+3+\frac{1}{x^6}\right)}.$$ Multiply numerator and denominator by $x^6$: $$u' = \frac{3x^2(x^6-1)}{x^{12}+3x^6+1}.$$ Now, $$3x^2(x^6-1)=3(x^8-x^2),$$ therefore $$u' = \frac{3(x^8-x^2)}{x^{12}+3x^6+1}.$$ 4. Substitute into the given integral: $$I=\int \frac{(x^8-x^2)\,dx}{(x^{12}+3x^6+1)\tan^{-1}\left(x^3+\frac{1}{x^3}\right)}.$$ Using the expression for $u'$, we get $$\frac{x^8-x^2}{x^{12}+3x^6+1}=\frac{1}{3}u'.$$ Thus, $$I=\frac{1}{3}\int \frac{u'}{u}\,dx.$$ 5. Integrate: $$I=\frac{1}{3}\ln|u|+C =\frac{1}{3}\ln\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|+C.$$ Using log property, $$\frac{1}{3}\ln|u|=\ln|u|^{1/3}.$$ So, $$I=\ln\left(\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|^{1/3}\right)+C.$$ 6. Compare with options: This matches **Option A**.

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View written solutionFree

Correct answer: A

Let $u=\tan^{-1}\left(x^3+\frac{1}{x^3}\right).$ Then the integral becomes of the form $\int \frac{u'}{u}\,dx,$ so we first compute $u'$ .
Differentiate the inner function: $\frac{d}{dx}\left(x^3+\frac{1}{x^3}\right)=3x^2-\frac{3}{x^4}=3\left(x^2-\frac{1}{x^4}\right)=\frac{3(x^6-1)}{x^4}.$

Hence, $u' = \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2}\cdot \frac{3(x^6-1)}{x^4}.$

= x^6+3+\frac{1}{x^6}.$$ So, $$u' = \frac{3(x^6-1)}{x^4\left(x^6+3+\frac{1}{x^6}\right)}.$$ Multiply numerator and denominator by $x^6$: $$u' = \frac{3x^2(x^6-1)}{x^{12}+3x^6+1}.$$ Now, $$3x^2(x^6-1)=3(x^8-x^2),$$ therefore $$u' = \frac{3(x^8-x^2)}{x^{12}+3x^6+1}.$$ 4. Substitute into the given integral: $$I=\int \frac{(x^8-x^2)\,dx}{(x^{12}+3x^6+1)\tan^{-1}\left(x^3+\frac{1}{x^3}\right)}.$$ Using the expression for $u'$, we get $$\frac{x^8-x^2}{x^{12}+3x^6+1}=\frac{1}{3}u'.$$ Thus, $$I=\frac{1}{3}\int \frac{u'}{u}\,dx.$$ 5. Integrate: $$I=\frac{1}{3}\ln|u|+C =\frac{1}{3}\ln\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|+C.$$ Using log property, $$\frac{1}{3}\ln|u|=\ln|u|^{1/3}.$$ So, $$I=\ln\left(\left|\tan^{-1}\left(x^3+\frac{1}{x^3}\right)\right|^{1/3}\right)+C.$$ 6. Compare with options: This matches **Option A**.