Question

integration of 1/x^2+3x+7​

Answer 1

Answer:

$\large\bold\red{\frac{2}{ \sqrt{19} } \: { \tan }^{ - 1} (\frac{2x + 3}{ \sqrt{19} }) + c}$

Step-by-step explanation:

Given,

We have to integrate ,

$\large{\int \frac{1}{ {x}^{2} + 3x + 7} dx}$

Further simplifying,

We get,

$= \int \frac{1}{ {(x)}^{2} + (2 \times \frac{3}{2} \times x) + {( \frac{3}{2} )}^{2} + 7 - {( \frac{3}{2} )}^{2} }dx \\ \\ = \int \frac{1}{ {(x + \frac{3}{2} )}^{2} + 7 - \frac{9}{4} } dx\\ \\ = \int \frac{1}{ {(x + \frac{3}{2} )}^{2} + \frac{28 - 9}{4} } dx \\ \\ = \int \frac{1}{ {(x + \frac{3}{2}) }^{2} + \frac{19}{4} } dx\\ \\ = \int \frac{1}{ {(x + \frac{3}{2} )}^{2} + {( \frac{ \sqrt{19} }{2} )}^{2} } dx$

Now,

Let's assume that,

$x + \frac{3}{2} = t$

Differentiating both the sides,

We get,

$= > dx = dt$

Substituting the values,

We get,

$= \int \frac{1}{ {t}^{2} + {( \frac{ \sqrt{19} }{2} )}^{2} } dt \\ \\ = \frac{1}{ \frac{ \sqrt{19} }{2} } \: { \tan }^{ - 1} (\frac{t}{ \frac{ \sqrt{19} }{2} } ) + c \\ \\ = \frac{2}{ \sqrt{19} } \: { \tan}^{ - 1} ( \frac{2t}{ \sqrt{19} } ) + c$

Substituting the value of 't',

We get,

$= \frac{2}{ \sqrt{19} } \: { \tan}^{ - 1} ( \frac{2(x + \frac{3}{2} )}{ \sqrt{19} } ) + c \\ \\ = \large \boxed{ \bold{\frac{2}{ \sqrt{19} } \: { \tan }^{ - 1} (\frac{2x + 3}{ \sqrt{19} }) + c}}$

Answer 2

Question we have to Find :---- integration of 1/(x^2+3x+7)

First we solve denominator by splitting the middle term ..

${x}^{2} + 3x + 7 \\ \\ here \: 3x = 2ab \\ where \: a = x \\ \\ so \: 2xb = 3x \\ b = \frac{3}{2} \\ \\ now \: adding \: and \: subtracting( \frac{3}{2} )^{2} we \: get \\ \\ {x}^{2} + 3x + 7 + (\frac{3}{2} )^{2} - (\frac{3}{2} )^{2} \: \\ \\ {x}^{2} + 3x + (\frac{3}{2} )^{2} + 7 - (\frac{3}{2} )^{2} \\ \\ (x + \frac{3}{2} )^{2} + (7 - \frac{9}{4} ) \\ \\ (x + \frac{3}{2} )^{2} + \frac{19}{4} \\ \\ (x + \frac{3}{2} )^{2} + ( \frac{ \sqrt{19} }{2} )^{2}$

Now see solution in image .....