CUET UG 2025 — Mathematics Calculus

The integral is $\displaystyle \int \frac{dx}{e^x + e^{-x}}$∫ex+e−xdx​

Multiply both numerator and denominator by $e^x$ex:

$$\int \frac{dx}{e^x + e^{-x}}$$∫ex+e−xdx​

$$= \int \frac{e^x \cdot dx}{e^x(e^x + e^{-x})}$$=∫ex(ex+e−x)ex⋅dx​

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Expand the denominator:

$$= \int \frac{e^x \, dx}{e^x \cdot e^x + e^x \cdot e^{-x}}$$=∫ex⋅ex+ex⋅e−xexdx​

$$= \int \frac{e^x \, dx}{e^{2x} + 1}$$=∫e2x+1exdx​

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Let $t = e^x$t=ex

Then $\frac{dt}{dx} = e^x$dxdt​=ex

So $dt = e^x \, dx$dt=exdx

Substituting:

$$\int \frac{e^x \, dx}{e^{2x} + 1}$$∫e2x+1exdx​

$$= \int \frac{dt}{t^2 + 1}$$=∫t2+1dt​

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Using the standard integral formula:

$$\int \frac{dt}{t^2 + 1} = \tan^{-1}(t) + c$$∫t2+1dt​=tan−1(t)+c

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Substituting back $t = e^x$t=ex:

$$\tan^{-1}(e^x) + c$$tan−1(ex)+c