$ \int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1} d x $ ની કિંમત શોધો.

  • A
    $ e^{x} \tan ^{-1} x+c $
  • B
    $ \tan ^{-1}\left(e^{x}\right)+c $
  • C
    $ \tan ^{-1}\left(x^{e}\right)+c $
  • D
    $ e^{\tan ^{-1} x}+c $

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$\int e^{x} \frac{(x-1)}{x^{2}} d x$ ની કિંમત શોધો.

$\int e^x(x+1)^2 dx=$

$\int e^x \tan x(1+\tan x) \, dx = $ . . . . . . $+ C$.

જો $\int \frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x)+c$ હોય,તો $f(x)$ શોધો.

$\int {\frac{{{e^x}(x - 1)}}{{{x^2}}}\;dx = } $

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