Mathematics

Evaluate : $$\displaystyle \int_{-1}^{1}\dfrac {1}{x^2 +2x+5}dx$$


SOLUTION
$$I=\displaystyle \int_{-1}^1\dfrac {1}{x^2+2x+5}dx $$ 

$$=\displaystyle \int_{-1}^1\dfrac {1}{x^2+2x+4+1}dx $$ 

$$=\displaystyle \int_{-1}^1\dfrac {1}{(x+1)^2 +2^2}dx$$

we know that $$=\displaystyle \int_{-1}^1\dfrac {1}{(x)^2 +a^2}dx=\dfrac {1}{2}\left [\tan^{-1} \dfrac {x}{a}\right]_{1-}^1 $$ 

$$=\dfrac {1}{a}\left [\tan^{-1} \dfrac {x+1}{2}\right]_{1-}^1 $$ 

$$=\dfrac {1}{2}(\tan^{-1} 1-\tan^{-1}0)=\dfrac{1}{2}\times \dfrac{\pi}{4}$$ 

$$=\dfrac {\pi}{8}$$

View Full Answer

Its FREE, you're just one step away


Subjective Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109
Enroll Now For FREE

Realted Questions

Q1 Single Correct Medium
$$\displaystyle \int \dfrac {\sin x \cos x}{\sqrt {1 - \sin^{4}x}}dx =$$
  • A. $$\dfrac {1}{2}\,\cos^{-1} (\sin^{2}x) + C$$
  • B. $$\tan^{-1}(\sin^{2} x) + C$$
  • C. $$\tan^{-1}(2\sin x) + C$$
  • D. $$\dfrac {1}{2}\, \sin^{-1}(\sin^{2}x) + C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Medium
Evaluate: $$\int _{ -1 }^{ 1 }{ { e }^{ x } } dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Medium
Integrate :
$$\displaystyle\int {\frac{cos4x+1}{cotx-tanx}dx}$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Integrate:
$$\displaystyle\int \dfrac { v } { 1 - v } =$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Medium
Integrate the following
i) $$\log{x}$$
ii) $${\cos}^{-1}(\sqrt{x})$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer