CUET UG 2025 — Mathematics Calculus

Given: $\displaystyle \int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}dx$∫e2x+e−2xe2x−e−2x​dx

The numerator is almost the derivative of the denominator.

Let $u = e^{2x} + e^{-2x}$u=e2x+e−2x

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Differentiate $u$u with respect to $x$x:

$$\frac{du}{dx} = \frac{d}{dx}(e^{2x}) + \frac{d}{dx}(e^{-2x})$$dxdu​=dxd​(e2x)+dxd​(e−2x)

$$\frac{du}{dx} = 2e^{2x} - 2e^{-2x}$$dxdu​=2e2x−2e−2x

$$\frac{du}{dx} = 2(e^{2x} - e^{-2x})$$dxdu​=2(e2x−e−2x)

Therefore: $du = 2(e^{2x} - e^{-2x})dx$du=2(e2x−e−2x)dx

$$(e^{2x} - e^{-2x})dx = \frac{1}{2}du$$(e2x−e−2x)dx=21​du

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$$\int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}dx = \int \frac{1}{u} \cdot \frac{1}{2}du$$∫e2x+e−2xe2x−e−2x​dx=∫u1​⋅21​du

$$= \frac{1}{2}\int \frac{1}{u}du$$=21​∫u1​du

$$= \frac{1}{2}\log_e|u| + C$$=21​loge​∣u∣+C

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Replace $u$u with $e^{2x} + e^{-2x}$e2x+e−2x:

$$= \frac{1}{2}\log_e|e^{2x} + e^{-2x}| + C$$=21​loge​∣e2x+e−2x∣+C

Therefore, $\displaystyle \int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}dx = \frac{1}{2}\log_e|e^{2x} + e^{-2x}| + C$∫e2x+e−2xe2x−e−2x​dx=21​loge​∣e2x+e−2x∣+C