Integration @ 02

Calculus Level 5

Y = ∫ 0 1 2 x 2 + 3 x + 3 ( x + 1 ) ( x 2 + 2 x + 2 ) d x Y = \int _{ 0 }^{ 1 }{ \frac { 2{ x }^{ 2 }+3x+3 }{ (x+1)({ x }^{ 2 }+2x+2) } } dx

Which of the following options are equal to Y ? Y?

(A) π 4 + 2 log ⁡ 2 − arctan ⁡ 2 \ \frac { \pi }{ 4 } +2\log2-\arctan { 2 }

(B) π 4 + 2 log ⁡ 2 − arctan ⁡ 1 3 \ \frac { \pi }{ 4 } +2\log2-\arctan {\frac{1}{3}}

(C) 2 log ⁡ 2 − arccot 3 \ 2 \log 2 - \text{arccot 3}

(D) − π 4 + log ⁡ 4 + arccot 2 \ -\frac{\pi}{4}+\log4+\text{arccot 2}

(A), (B) and (C) (A), (B), (C) and (D) (A), (B) and (D) (A) and (C) (A), (C) and (D) (A) and (D) (B) and (C) (B), (C) and (D)

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Saurav Pal by Saurav Pal , 26, India

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2 solutions

Akhilesh Vibhute
Nov 28, 2015

LOL! No need to integrate just Simplify the options and you will get option A, C and D as same..... So this is how we make life absolutely easy :-) As simple as that...

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Brilliant is for studying and for that you need to try solve the questions.....No one cares if you cheat.

Anandhu Raj - 5 years, 3 months ago

Why do that when integrating is much more easier.

Kunal Verma - 5 years, 2 months ago

Y = ∫ 0 1 2 x 2 + 3 x + 3 ( x + 1 ) ( x 2 + 2 x + 2 ) d x By partial fraction decomposition = ∫ 0 1 ( 2 x + 1 − 1 x 2 + 2 x + 2 ) d x = ∫ 0 1 ( 2 x + 1 − 1 ( x + 1 ) 2 + 1 ) d x = 2 log ⁡ ( x + 1 ) − arctan ⁡ ( x + 1 ) ∣ 0 1 = 2 log ⁡ 2 − arctan ⁡ 2 + π 4 . . . ( A ) \begin{aligned} Y & = \int_0^1 \frac {2x^2+3x+3}{(x+1)(x^2+2x+2)} dx & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \int_0^1 \left(\frac 2{x+1} - \frac 1{x^2+2x+2} \right) dx \\ & = \int_0^1 \left(\frac 2{x+1} - \frac 1{(x+1)^2+1} \right) dx \\ & = 2 \log (x+1) - \arctan (x+1) \bigg|_0^1 \\ & = 2\log 2 \ {\color{#3D99F6} - \arctan 2 + \frac \pi 4} \quad ...(A) \end{aligned}

From (A):

A = 2 log ⁡ 2 + π 4 − arctan ⁡ 2 = 2 log ⁡ 2 + arctan ⁡ ( 1 − 2 1 + 2 ) = 2 log ⁡ 2 − arccot 3 . . . ( C ) \begin{aligned} A & = 2\log 2 + \color{#3D99F6} \frac \pi 4 - \arctan 2 \\ & = 2\log 2 + \color{#3D99F6} \arctan \left( \frac {1-2}{1+2} \right) \\ & = 2\log 2 \ \color{#3D99F6} - \text{arccot 3} & ...(C) \end{aligned}

From (A):

A = 2 log ⁡ 2 + π 4 − arctan ⁡ 2 = − π 4 + log ⁡ 4 + π 2 − arctan ⁡ 2 = − π 4 + log ⁡ 4 + arccot 2 . . . ( D ) \begin{aligned} A & = 2\log 2 + \frac \pi 4 - \arctan 2 \\ & = - \frac \pi 4 + \log 4 + \color{#3D99F6} \frac \pi 2 - \arctan 2 \\ & = - \frac \pi 4 + \log 4 + \color{#3D99F6} \text{arccot 2} &...(D) \end{aligned}

Therefore, Y Y is equal to options (A), (C) and (D) .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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