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Definite Integration question

2025 · 2 Apr · Shift 2 · Q43
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Definite Integration question

2025 · 2 Apr · Shift 2 · Q43

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
4∫01(13+x2+1+x2)dx−3log⁡e(3)4 \int_0^1\left(\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right) d x-3 \log _e(\sqrt{3})4∫01​(3+x2​+1+x2​1​)dx−3loge​(3​) is equal to :
  1. A
    2−2−log⁡e(1+2)2-\sqrt{2}-\log _{\mathrm{e}}(1+\sqrt{2})2−2​−loge​(1+2​)
  2. B
    2+2+log⁡e(1+2)2+\sqrt{2}+\log _{\mathrm{e}}(1+\sqrt{2})2+2​+loge​(1+2​)
  3. C
    2+2−log⁡e(1+2)2+\sqrt{2}-\log _{\mathrm{e}}(1+\sqrt{2})2+2​−loge​(1+2​)
  4. D
    2−2+log⁡e(1+2)2-\sqrt{2}+\log _e(1+\sqrt{2})2−2​+loge​(1+2​)
View written solutionFree

Correct answer: A

  1. Simplify the integrand

We need to evaluate 4∫01dx3+x2+1+x2−3ln⁡(3).4\int_0^1 \frac{dx}{\sqrt{3+x^2}+\sqrt{1+x^2}}-3\ln(\sqrt{3}).4∫01​3+x2​+1+x2​dx​−3ln(3​).

Rationalize the denominator inside the integral: 13+x2+1+x2⋅3+x2−1+x23+x2−1+x2\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\cdot \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{\sqrt{3+x^2}-\sqrt{1+x^2}}3+x2​+1+x2​1​⋅3+x2​−1+x2​3+x2​−1+x2​​

Using (3+x2)−(1+x2)=2,(3+x^2)-(1+x^2)=2,(3+x2)−(1+x2)=2, we get 13+x2+1+x2=3+x2−1+x22.\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}=\frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{2}.3+x2​+1+x2​1​=23+x2​−1+x2​​.

Hence, 4∫01dx3+x2+1+x2=2∫01(x2+3−x2+1)dx.4\int_0^1 \frac{dx}{\sqrt{3+x^2}+\sqrt{1+x^2}}=2\int_0^1 \left(\sqrt{x^2+3}-\sqrt{x^2+1}\right)dx.4∫01​3+x2​+1+x2​dx​=2∫01​(x2+3​−x2+1​)dx.

So the expression becomes 2∫01x2+3 dx−2∫01x2+1 dx−3ln⁡(3).2\int_0^1 \sqrt{x^2+3}\,dx-2\int_0^1 \sqrt{x^2+1}\,dx-3\ln(\sqrt{3}).2∫01​x2+3​dx−2∫01​x2+1​dx−3ln(3​).

  1. Use the standard integral

Recall: ∫x2+a2 dx=x2x2+a2+a22ln⁡(x+x2+a2)+C.\int \sqrt{x^2+a^2}\,dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln\left(x+\sqrt{x^2+a^2}\right)+C.∫x2+a2​dx=2x​x2+a2​+2a2​ln(x+x2+a2​)+C.

  1. Evaluate ∫01x2+3 dx\int_0^1 \sqrt{x^2+3}\,dx∫01​x2+3​dx

Here a2=3a^2=3a2=3: ∫x2+3 dx=x2x2+3+32ln⁡(x+x2+3).\int \sqrt{x^2+3}\,dx=\frac{x}{2}\sqrt{x^2+3}+\frac{3}{2}\ln\left(x+\sqrt{x^2+3}\right).∫x2+3​dx=2x​x2+3​+23​ln(x+x2+3​).

Thus,

\left[\frac{x}{2}\sqrt{x^2+3}+\frac{3}{2}\ln\left(x+\sqrt{x^2+3}\right)\right]_0^1.$$ At $x=1$: $$\frac{1}{2}\sqrt{4}+\frac{3}{2}\ln(1+2)=1+\frac{3}{2}\ln 3.$$ At $x=0$: $$0+\frac{3}{2}\ln(\sqrt{3}).$$ So, $$\int_0^1 \sqrt{x^2+3}\,dx=1+\frac{3}{2}\ln 3-\frac{3}{2}\ln(\sqrt{3}).$$ Since $$\ln(\sqrt{3})=\frac{1}{2}\ln 3,$$ this becomes $$1+\frac{3}{2}\ln 3-\frac{3}{4}\ln 3=1+\frac{3}{4}\ln 3.$$ Therefore, $$2\int_0^1 \sqrt{x^2+3}\,dx=2+\frac{3}{2}\ln 3.$$ --- 4. **Evaluate** $\int_0^1 \sqrt{x^2+1}\,dx$ Here $a^2=1$: $$\int \sqrt{x^2+1}\,dx=\frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\ln\left(x+\sqrt{x^2+1}\right).$$ Thus, $$\int_0^1 \sqrt{x^2+1}\,dx= \left[\frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\ln\left(x+\sqrt{x^2+1}\right)\right]_0^1.$$ At $x=1$: $$\frac{\sqrt{2}}{2}+\frac{1}{2}\ln(1+\sqrt{2}).$$ At $x=0$: $$0+\frac{1}{2}\ln 1=0.$$ So, $$\int_0^1 \sqrt{x^2+1}\,dx=\frac{\sqrt{2}}{2}+\frac{1}{2}\ln(1+\sqrt{2}).$$ Therefore, $$2\int_0^1 \sqrt{x^2+1}\,dx=\sqrt{2}+\ln(1+\sqrt{2}).$$ --- 5. **Combine all parts** The required expression is $$\left(2+\frac{3}{2}\ln 3\right)-\left(\sqrt{2}+\ln(1+\sqrt{2})\right)-3\ln(\sqrt{3}).$$ Again, $$3\ln(\sqrt{3})=3\cdot \frac{1}{2}\ln 3=\frac{3}{2}\ln 3.$$ So the logarithmic terms in $\ln 3$ cancel: $$2+\frac{3}{2}\ln 3-\sqrt{2}-\ln(1+\sqrt{2})-\frac{3}{2}\ln 3$$ $$=2-\sqrt{2}-\ln(1+\sqrt{2}).$$ --- 6. **Match with options** This is exactly: $$2-\sqrt{2}-\log_e(1+\sqrt{2}).$$ So the correct option is **A**.
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