Q.20. The integral is equal to :

Q: What is the correct answer for JEE Main Mathematics Q20?

The correct answer is D. Sol. $\int_0^1 \cot^{-1} (1 + x + x^2) \, dx = \int_0^1 \tan^{-1} \left(\frac{1}{1+x(x+1)}\right) \, dx$ $= \int_0^1 \tan^{-1} \left(\frac{(x+1)-x}{1+x(x+1)}\right) \, dx$ $= \int_0^1 (\tan^{-1}(x+1) - \tan^{-1}x) \, dx$ $= [x \tan^{-1}(x+1)]_0^1 - \int_0^1 \frac{x}{1+(x+1)^2} \, dx - [x \tan^{-1}x]_0^1 + \int_0^1 \frac{x}{1+x^2} \, dx$ $= (1 \cdot \tan^{-1}2 - 0) - \int_0^1 \frac{x}{x^2+2x+2} \,

Q: Q.20. The integral is equal to :

The correct answer is D. Sol. $\int_0^1 \cot^{-1} (1 + x + x^2) \, dx = \int_0^1 \tan^{-1} \left(\frac{1}{1+x(x+1)}\right) \, dx$ $= \int_0^1 \tan^{-1} \left(\frac{(x+1)-x}{1+x(x+1)}\right) \, dx$ $= \int_0^1 (\tan^{-1}(x+1) - \tan^{-1}x) \, dx$ $= [x \tan^{-1}(x+1)]_0^1 - \int_0^1 \frac{x}{1+(x+1)^2} \, dx - [x \tan^{-1}x]_0^1 + \int_0^1 \frac{x}{1+x^2} \, dx$ $= (1 \cdot \tan^{-1}2 - 0) - \int_0^1 \frac{x}{x^2+2x+2} \,

Q: What are the options for this JEE Main Mathematics question?

A. $2\tan^{-1}2 + \frac{1}{2}\log \frac{5}{4} - \frac{\pi}{2}$ · B. $2\tan^{-1}2 + \log \frac{5}{4} - \frac{\pi}{2}$ · C. $2\tan^{-1}2 - \frac{1}{2}\log \frac{5}{4} - \frac{\pi}{2}$ · D. $2\tan^{-1}2 - \log \frac{5}{4} - \frac{\pi}{2}$

Question

Q.20. The integral  is equal to :

FirstInTest · Accepted Answer

Sol. $\int_0^1 \cot^{-1} (1 + x + x^2) \, dx = \int_0^1 	an^{-1} \left(\frac{1}{1+x(x+1)}ight) \, dx$
$= \int_0^1 	an^{-1} \left(\frac{(x+1)-x}{1+x(x+1)}ight) \, dx$
$= \int_0^1 (	an^{-1}(x+1) - 	an^{-1}x) \, dx$
$= [x 	an^{-1}(x+1)]_0^1 - \int_0^1 \frac{x}{1+(x+1)^2} \, dx - [x 	an^{-1}x]_0^1 + \int_0^1 \frac{x}{1+x^2} \, dx$
$= (1 \cdot 	an^{-1}2 - 0) - \int_0^1 \frac{x}{x^2+2x+2} \, dx - (1 \cdot 	an^{-1}1 - 0) + \int_0^1 \frac{x}{1+x^2} \, dx$
$= 	an^{-1}2 - \frac{\pi}{4} - \frac{1}{2} \int_0^1 \frac{2x+2-2}{x^2+2x+2} \, dx + \frac{1}{2} \int_0^1 \frac{2x}{1+x^2} \, dx$
$= 	an^{-1}2 - \frac{\pi}{4} - \frac{1}{2} [\ln(x^2+2x+2)]_0^1 + \int_0^1 \frac{1}{(x+1)^2+1} \, dx + \frac{1}{2} [\ln(1+x^2)]_0^1$
$= 	an^{-1}2 - \frac{\pi}{4} - \frac{1}{2} (\ln5 - \ln2) + [	an^{-1}(x+1)]_0^1 + \frac{1}{2} (\ln2 - \ln1)$
$= 	an^{-1}2 - \frac{\pi}{4} - \frac{1}{2} \ln(5/2) + (	an^{-1}2 - 	an^{-1}1) + \frac{1}{2} \ln2$
$= 	an^{-1}2 - \frac{\pi}{4} - \frac{1}{2} \ln5 + \frac{1}{2} \ln2 + 	an^{-1}2 - \frac{\pi}{4} + \frac{1}{2} \ln2$
$= 2	an^{-1}2 - \frac{\pi}{2} + \ln2 - \frac{1}{2} \ln5 = 2	an^{-1}2 - \frac{\pi}{2} + \ln(\frac{2}{\sqrt{5}})$

The solution provided in the image is:
Sol. = $\int_0^1 	an^{-1} \left(\frac{1}{1+x+x^2}ight)dx$
= $\int_0^1 	an^{-1} \left(\frac{(x+1)-x}{1+x(x+1)}ight)dx$
= $\int_0^1 (	an^{-1}(x+1) - 	an^{-1}x)dx$
= $[x 	an^{-1}(x+1)]_0^1 - \int_0^1 \frac{x}{1+(x+1)^2}dx - [x 	an^{-1}x]_0^1 + \int_0^1 \frac{x}{1+x^2}dx$
= $(1 	an^{-1}2 - 0) - \left[\frac{1}{2} \ln(1+(x+1)^2) - 	an^{-1}(x+1)ight]_0^1 - (1 	an^{-1}1 - 0) + \left[\frac{1}{2} \ln(1+x^2)ight]_0^1$
= $	an^{-1}2 - \left(\frac{1}{2} \ln5 - 	an^{-1}2ight) + \left(\frac{1}{2} \ln2 - 	an^{-1}1ight) - 	an^{-1}1 + \frac{1}{2} \ln2$
= $	an^{-1}2 - \frac{1}{2} \ln5 + 	an^{-1}2 + \frac{1}{2} \ln2 - \frac{\pi}{4} - \frac{\pi}{4} + \frac{1}{2} \ln2$
= $2	an^{-1}2 - \frac{\pi}{2} + \ln2 - \frac{1}{2} \ln5 = 2	an^{-1}2 - \frac{\pi}{2} + \ln(\frac{2}{\sqrt{5}})$ 
This does not match any option directly. Let's re-check the options and the solution steps. The image's solution has a typo in the second line of the final calculation. It should be $\frac{1}{2} \ln(1+x^2)$ not $\frac{1}{2} \ln(1+x^2)$.

Let's re-evaluate the definite integral $\int_0^1 (	an^{-1}(x+1) - 	an^{-1}x)dx$
Using integration by parts: $\int u dv = uv - \int v du$
Let $u = 	an^{-1}(x+1)$, $dv = dx \Rightarrow du = \frac{1}{1+(x+1)^2}dx$, $v = x$
$\int_0^1 	an^{-1}(x+1)dx = [x 	an^{-1}(x+1)]_0^1 - \int_0^1 \frac{x}{1+(x+1)^2}dx$
$= (1 \cdot 	an^{-1}2 - 0) - \int_0^1 \frac{x}{x^2+2x+2}dx$
$= 	an^{-1}2 - \frac{1}{2} \int_0^1 \frac{2x+2-2}{x^2+2x+2}dx = 	an^{-1}2 - \frac{1}{2} [\ln(x^2+2x+2)]_0^1 + \int_0^1 \frac{1}{(x+1)^2+1}dx$
$= 	an^{-1}2 - \frac{1}{2}(\ln5 - \ln2) + [	an^{-1}(x+1)]_0^1$
$= 	an^{-1}2 - \frac{1}{2}\ln(5/2) + (	an^{-1}2 - 	an^{-1}1) = 2	an^{-1}2 - \frac{\pi}{4} - \frac{1}{2}\ln(5/2)$

Let $u = 	an^{-1}x$, $dv = dx \Rightarrow du = \frac{1}{1+x^2}dx$, $v = x$
$\int_0^1 	an^{-1}x dx = [x 	an^{-1}x]_0^1 - \int_0^1 \frac{x}{1+x^2}dx$
$= (1 \cdot 	an^{-1}1 - 0) - \frac{1}{2} [\ln(1+x^2)]_0^1 = \frac{\pi}{4} - \frac{1}{2}(\ln2 - \ln1) = \frac{\pi}{4} - \frac{1}{2}\ln2$

So, $\int_0^1 (	an^{-1}(x+1) - 	an^{-1}x)dx = (2	an^{-1}2 - \frac{\pi}{4} - \frac{1}{2}\ln(5/2)) - (\frac{\pi}{4} - \frac{1}{2}\ln2)$
$= 2	an^{-1}2 - \frac{\pi}{2} - \frac{1}{2}\ln5 + \frac{1}{2}\ln2 + \frac{1}{2}\ln2 = 2	an^{-1}2 - \frac{\pi}{2} + \ln2 - \frac{1}{2}\ln5$
$= 2	an^{-1}2 - \frac{\pi}{2} + \ln(\frac{2}{\sqrt{5}})$ 
This is still not matching the options. Let's re-check the options and the image's solution. The image's solution has a final result of $2	an^{-1}2 - \log \frac{5}{4} - \frac{\pi}{2}$. This matches option (4) if $\log$ means $\ln$. 
$2	an^{-1}2 - \ln(5/4) - \frac{\pi}{2} = 2	an^{-1}2 - (\ln5 - \ln4) - \frac{\pi}{2} = 2	an^{-1}2 - \ln5 + 2\ln2 - \frac{\pi}{2}$.
My calculation was $2	an^{-1}2 - \frac{\pi}{2} + \ln2 - \frac{1}{2}\ln5$. These are different.

Let's re-examine the image's solution steps carefully. The image's solution has a step: 
$= (2	an^{-1}2 - \frac{1}{2}\ln2 - \frac{\pi}{4}) - (\frac{1}{2}\ln2 - \frac{\pi}{4})$
This step is incorrect. The previous line was: 
$= (2	an^{-1}2 - \frac{1}{2}\ln5 + \frac{1}{2}\ln2 - \frac{\pi}{4}) - (\frac{\pi}{4} - \frac{1}{2}\ln2)$ 
$= 2	an^{-1}2 - \frac{1}{2}\ln5 + \frac{1}{2}\ln2 - \frac{\pi}{4} - \frac{\pi}{4} + \frac{1}{2}\ln2$
$= 2	an^{-1}2 - \frac{\pi}{2} + \ln2 - \frac{1}{2}\ln5$. This is what I got.

However, the image's final answer is $2	an^{-1}2 - \log \frac{5}{4} - \frac{\pi}{2}$. This implies that $\ln2 - \frac{1}{2}\ln5 = -\ln(5/4)$.
$\ln2 - \frac{1}{2}\ln5 = \ln2 - \ln\sqrt{5} = \ln(\frac{2}{\sqrt{5}})$.
$-\ln(5/4) = \ln(4/5)$. These are not equal.

There is a discrepancy in the provided solution steps and the final answer. Given the options, option D is the closest if there's a typo in the solution's derivation. I will transcribe the solution as it is presented in the image, acknowledging the potential error.

Answer

$2	an^{-1}2 + \frac{1}{2}\log \frac{5}{4} - \frac{\pi}{2}$

Answer

$2	an^{-1}2 + \log \frac{5}{4} - \frac{\pi}{2}$

Answer

$2	an^{-1}2 - \frac{1}{2}\log \frac{5}{4} - \frac{\pi}{2}$

Answer

$2	an^{-1}2 - \log \frac{5}{4} - \frac{\pi}{2}$

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