Mathematics

# Evaluate : $\displaystyle \int { \dfrac { 1 }{ { 2x }^{ 2 }+x+1 } } dx$

##### SOLUTION
Let  $I=\displaystyle\int { \dfrac { 1 }{ { 2x }^{ 2 }+x+1 } } dx$

$=\dfrac{1}{2}\displaystyle\int { \dfrac { 1 }{ { x }^{ 2 }+\dfrac{x}{2}+\dfrac{1}{2} } } dx$

$=\dfrac{1}{2}\displaystyle\int { \dfrac { 1 }{ { x }^{ 2 }+\dfrac{x}{2}+\dfrac{1}{16}-\dfrac{1}{16}+\dfrac{1}{2} } } dx$

$=\dfrac{1}{2}\displaystyle\int { \dfrac {1} {{\left( {x+\dfrac{1}{4}}\right)^{2}}-\dfrac{1}{16}+\dfrac{1}{2}}}dx$

$=\dfrac{1}{2}\displaystyle\int { \dfrac {1} {{\left( {x+\dfrac{1}{4}}\right)^{2}}+\dfrac{7}{16}}}dx$

We know that,  $\displaystyle \int \dfrac {dx}{ {x^2+a^2} }=\dfrac{1}{a}\tan^{-1}\dfrac xa+C$

$=\dfrac{1}{2}\left(\dfrac{1}{{\dfrac{\sqrt7}{4}}}{tan}^{-1}\left[\dfrac{x+\dfrac{1}{4}}{\dfrac{\sqrt7}{4}}\right]\right)+c$

$I=\dfrac{2}{\sqrt7}{tan}^{-1}\left[\dfrac{(4x+1)}{\sqrt7}\right]+c$

Hence,

$\displaystyle\int { \dfrac { 1 }{ { 2x }^{ 2 }+x+1 } } dx$$=\dfrac{2}{\sqrt7}{tan}^{-1}\left[\dfrac{(4x+1)}{\sqrt7}\right]+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

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For $n\in N$, $0 < t < \pi/2$; the value of $\displaystyle\int^{n\pi +t}_0(|\cos x|+|\sin x|)dx=$?
• A. $4n+\sin t-\cos t +1$
• B. $4n-\sin t-\cos t+1$
• C. $2n-\sin t-\cos t+1$
• D. $2n-\sin t-\cos t-1$

1 Verified Answer | Published on 17th 09, 2020

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