Column I, a) ∫e2x+1e2x−1dx is equal to b) ∫(ex+e−x)21dx is equal to c) ∫1+exe−xdx is equal to d) ∫1−e2x1dx is equal to COLUMN II p) x−log[1+1−e2x+c q) log(ex+1)−x−e−x+c r) log(e2x+1)−x+c s) −2(e2x+1)1+c

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Column I, a) {{{e}^{{{2}{x}}}-{1}}}{{{e}^{{{2}{x}}}+{1}}}{.{d}

Question

Column I, a) $\int \frac{e ^{2 x} - 1}{e ^{2 x} + 1} d x$ is equal to b) $\int \frac{1}{( e ^{x} + e ^{- x} ) ^{2}} d x$ is equal to c) $\int \frac{e ^{- x}}{1 + e ^{x}} d x$ is equal to d) $\int \frac{1}{1 - e ^{2 x}} d x$ is equal to COLUMN II p) $x - lo g [1 + 1 - e^{2 x} + c$ q) $lo g (e^{x} + 1) - x - e^{- x} + c$ r) $lo g (e^{2 x} + 1) - x + c$ s) $- \frac{1}{2 ( e ^{2 x} + 1 )} + c$

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Samuel Discussed

Column I, a)

\int \frac{e ^{2 x} - 1}{e ^{2 x} + 1} d x

is equal to b)

\int \frac{1}{( e ^{x} + e ^{- x} ) ^{2}} d x

is equal to c)

\int \frac{e ^{- x}}{1 + e ^{x}} d x

is equal to d)

\int \frac{1}{1 - e ^{2 x}} d x

is equal to COLUMN II p)

x - lo g [1 + 1 - e^{2 x} + c

lo g (e^{x} + 1) - x - e^{- x} + c

lo g (e^{2 x} + 1) - x + c

- \frac{1}{2 ( e ^{2 x} + 1 )} + c

12 mins ago

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