Question

integral DX upon X square + 2 X + 2 ​

Answer 1

Answer:

$\large \bold\red{{ \tan }^{ - 1} (x + 1) + c}$

Step-by-step explanation:

We have to integrate,

$\displaystyle \int \frac{dx}{ {x}^{2} + 2x + 2 } \\ \\ \\ = \displaystyle \int \frac{dx}{( {x}^{2} + 2x + 1) + 1 } \\ \\ \\ = \displaystyle \int \frac{dx}{ {(x + 1)}^{2} + {(1)}^{2} }$

Now,

Let's assume that,

$x + 1 = t$

Differentiating both the sides,

We get,

$= > dx = dt$

Substituting the values,

We get,

$= \displaystyle \int \frac{dt}{ {t}^{2} + {1}^{2} } \\ \\ \\ = \displaystyle \int \frac{dt}{1 + {t}^{2} } \\ \\\\ = { \tan}^{ - 1} t + c$

Substituting the value of t,

We get,

$= \large \bold{{ \tan }^{ - 1} (x + 1) + c}$

where,

c is an integral constant.