MCQ
$\int \frac{1}{x^{2}+2 x+2} d x $ बराबर है:
  • A
    tan-1 x + C
  • B
    (x + 1) tan-1 x + C
  • C
    x tan-1 (x + 1) + C
  • D
    tan-1 (x + 1) + C

Answer

माना $I=\int \frac{1}{x^{2}+2 x+2} d x$ $=\int \frac{1}{(x+1)^{2}+1^{2}} d x$
माना x + 1 = t $\Rightarrow$ dx = dt
$\therefore$ $I=\int \frac{1}{t^{2}+1^{2}} d t$ $=\frac{1}{1} \tan ^{-1}\left(\frac{t}{1}\right)$ + C [$\because$ $\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)$]
$=\tan ^{-1}\left(\frac{x+1}{1}\right)+C$ = tan−1 (x + 1) + C

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