The value of the integral \int\limits_ - 2^2 |x^3 + x| \over (e^x|x| + 1)dx is equal to :

Definite Integration PYQ — JEE Main 2022 · 27 Jun · Shift 1 · Q27 with Solution

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Correct answer: D

We need to evaluate $I=\int_{-2}^{2} \frac{|x^3+x|}{e^{x|x|}+1}\,dx.$
First simplify the numerator: $x^3+x=x(x^2+1).$ Since $x^2+1>0$ for all $x$ , we get $|x^3+x|=|x|(x^2+1).$ So, $I=\int_{-2}^{2} \frac{|x|(x^2+1)}{e^{x|x|}+1}\,dx.$
Now split the integral over $[-2,0]$ and $[0,2]$ .

For $x\ge 0$ : $|x|=x,\qquad x|x|=x^2.$ Hence, $\frac{|x|(x^2+1)}{e^{x|x|}+1}=\frac{x(x^2+1)}{e^{x^2}+1}.$

For $x<0$ : $|x|=-x,\qquad x|x|=-x^2.$ Hence, $\frac{|x|(x^2+1)}{e^{x|x|}+1}=\frac{(-x)(x^2+1)}{e^{-x^2}+1}.$

Thus, $I=\int_{-2}^{0}\frac{(-x)(x^2+1)}{e^{-x^2}+1}\,dx+\int_{0}^{2}\frac{x(x^2+1)}{e^{x^2}+1}\,dx.$

In the first integral, put $x=-t$ . Then $dx=-dt$ , and when $x=-2$ , $t=2$ ; when $x=0$ , $t=0$ . So,

=\int_{2}^{0}\frac{t(t^2+1)}{e^{-t^2}+1}(-dt) =\int_{0}^{2}\frac{t(t^2+1)}{e^{-t^2}+1}\,dt.$$ Therefore, $$I=\int_{0}^{2} t(t^2+1)\left(\frac{1}{e^{-t^2}+1}+\frac{1}{e^{t^2}+1}\right)dt.$$ 5. Use the identity $$\frac{1}{1+e^{-u}}+\frac{1}{1+e^{u}}=1.$$ Indeed, $$\frac{1}{1+e^{-u}}=\frac{e^u}{1+e^u},$$ so $$\frac{e^u}{1+e^u}+\frac{1}{1+e^u}=1.$$ With $u=t^2$, we get $$\frac{1}{e^{-t^2}+1}+\frac{1}{e^{t^2}+1}=1.$$ Hence, $$I=\int_{0}^{2} t(t^2+1)\,dt.$$ 6. Now evaluate: $$\int_{0}^{2} t(t^2+1)\,dt=\int_{0}^{2}(t^3+t)\,dt =\left[\frac{t^4}{4}+\frac{t^2}{2}\right]_0^2 =\frac{16}{4}+\frac{4}{2}=4+2=6.$$ 7. Therefore, $$\boxed{I=6}. $$ 8. Comparing with the options, this is **Option D**. 9. Comparison with stored correct answer: Stored correct answer is **D**, which matches our result.

View written solutionFree

Correct answer: D

We need to evaluate $I=\int_{-2}^{2} \frac{|x^3+x|}{e^{x|x|}+1}\,dx.$
First simplify the numerator: $x^3+x=x(x^2+1).$ Since $x^2+1>0$ for all $x$ , we get $|x^3+x|=|x|(x^2+1).$ So, $I=\int_{-2}^{2} \frac{|x|(x^2+1)}{e^{x|x|}+1}\,dx.$
Now split the integral over $[-2,0]$ and $[0,2]$ .

For $x\ge 0$ : $|x|=x,\qquad x|x|=x^2.$ Hence, $\frac{|x|(x^2+1)}{e^{x|x|}+1}=\frac{x(x^2+1)}{e^{x^2}+1}.$

For $x<0$ : $|x|=-x,\qquad x|x|=-x^2.$ Hence, $\frac{|x|(x^2+1)}{e^{x|x|}+1}=\frac{(-x)(x^2+1)}{e^{-x^2}+1}.$

Thus, $I=\int_{-2}^{0}\frac{(-x)(x^2+1)}{e^{-x^2}+1}\,dx+\int_{0}^{2}\frac{x(x^2+1)}{e^{x^2}+1}\,dx.$

In the first integral, put $x=-t$ . Then $dx=-dt$ , and when $x=-2$ , $t=2$ ; when $x=0$ , $t=0$ . So,

=\int_{2}^{0}\frac{t(t^2+1)}{e^{-t^2}+1}(-dt) =\int_{0}^{2}\frac{t(t^2+1)}{e^{-t^2}+1}\,dt.$$ Therefore, $$I=\int_{0}^{2} t(t^2+1)\left(\frac{1}{e^{-t^2}+1}+\frac{1}{e^{t^2}+1}\right)dt.$$ 5. Use the identity $$\frac{1}{1+e^{-u}}+\frac{1}{1+e^{u}}=1.$$ Indeed, $$\frac{1}{1+e^{-u}}=\frac{e^u}{1+e^u},$$ so $$\frac{e^u}{1+e^u}+\frac{1}{1+e^u}=1.$$ With $u=t^2$, we get $$\frac{1}{e^{-t^2}+1}+\frac{1}{e^{t^2}+1}=1.$$ Hence, $$I=\int_{0}^{2} t(t^2+1)\,dt.$$ 6. Now evaluate: $$\int_{0}^{2} t(t^2+1)\,dt=\int_{0}^{2}(t^3+t)\,dt =\left[\frac{t^4}{4}+\frac{t^2}{2}\right]_0^2 =\frac{16}{4}+\frac{4}{2}=4+2=6.$$ 7. Therefore, $$\boxed{I=6}. $$ 8. Comparing with the options, this is **Option D**. 9. Comparison with stored correct answer: Stored correct answer is **D**, which matches our result.