Mathematics

# Solve:$\int\dfrac{x^{2}+1}{x^{2}-4x+6}dx$

##### SOLUTION
$\begin{array}{l} \int _{ }^{ }{ \dfrac { { { x^{ 2 } }+1 } }{ { { x^{ 2 } }-4x+6 } } dx } \\\\=\int _{ }^{ }{ \left( { \dfrac { { 4x-5 } }{ { { x^{ 2 } }-4x+6 } } +1 } \right) dx } \\\\ =\int _{ }^{ }{ \dfrac { { 4x-5 } }{ { { x^{ 2 } }-4x+6 } } dx } +\int _{ }^{ }{ 1dx } \\ =2\ln { \left( { { x^{ 2 } }-4x+6 } \right) } +x+\dfrac { { 3ta{ n^{ -1 } }\left( { \dfrac { { x-2 } }{ { \sqrt { 2 } } } } \right) } }{ { \sqrt { 2 } } } +C \end{array}$

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int_{0}^{\infty} \dfrac {x \ln x}{(1 + x^{2})^{2}}dx =$
• A. $1$
• B. $-1$
• C. $\dfrac {\pi}{2}$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The value of $\displaystyle \int(3x^{2}\tan\frac{1}{x}-x\sec^{2}\frac{1}{x})dx {\it}$ is
• A. $x^{2}\displaystyle \tan\frac{1}{x}+c$
• B. $x\displaystyle \tan\frac{1}{x}+c$
• C. $\displaystyle \tan\frac{1}{x}+c$
• D. $x^{3}\displaystyle \tan\frac{1}{x}+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
$I= \int \frac{1}{12+13cosx}dx.$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium

$\displaystyle \int_{1}^{\infty}\left( \frac { 1 }{ 1+x^{ 2 } } \right) d{ x }=$

• A. $-\displaystyle \frac{\pi}{4}$
• B. $\displaystyle \frac{\pi}{2}$
• C. $-\displaystyle \frac{\pi}{2}$
• D. $\displaystyle \frac{\pi}{4}$

$(x^2+1) \log x$