CUET UG 2025 — Mathematics Calculus

The integral contains $\sqrt{x+1}$x+1​ in the denominator. To simplify, let:

$$\sqrt{x+1} = t$$x+1​=t

$$x + 1 = t^2$$x+1=t2

$$x = t^2 - 1$$x=t2−1

$$dx = 2t \, dt$$dx=2tdt

The term $(x-3)$(x−3) becomes:

$$x - 3 = t^2 - 1 - 3 = t^2 - 4$$x−3=t2−1−3=t2−4

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Substituting into the integral:

$$\int \frac{2}{(x-3)\sqrt{x+1}} dx$$∫(x−3)x+1​2​dx

$$= \int \frac{2}{(t^2-4) \cdot t} \cdot 2t \, dt$$=∫(t2−4)⋅t2​⋅2tdt

$$= \int \frac{4t}{t(t^2-4)} dt$$=∫t(t2−4)4t​dt

$$= \int \frac{4}{t^2-4} dt$$=∫t2−44​dt

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The denominator factors as:

$$t^2 - 4 = (t-2)(t+2)$$t2−4=(t−2)(t+2)

Using partial fractions:

$$\frac{4}{(t-2)(t+2)} = \frac{A}{t-2} + \frac{B}{t+2}$$(t−2)(t+2)4​=t−2A​+t+2B​

$$4 = A(t+2) + B(t-2)$$4=A(t+2)+B(t−2)

For $t = 2$t=2:

$$4 = A(4)$$4=A(4)

$$A = 1$$A=1

For $t = -2$t=−2:

$$4 = B(-4)$$4=B(−4)

$$B = -1$$B=−1

Therefore:

$$\frac{4}{(t-2)(t+2)} = \frac{1}{t-2} - \frac{1}{t+2}$$(t−2)(t+2)4​=t−21​−t+21​

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Integrating:

$$\int \left(\frac{1}{t-2} - \frac{1}{t+2}\right) dt$$∫(t−21​−t+21​)dt

$$= \ln|t-2| - \ln|t+2| + C$$=ln∣t−2∣−ln∣t+2∣+C

$$= \ln\left|\frac{t-2}{t+2}\right| + C$$=ln​t+2t−2​​+C

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Substituting $t = \sqrt{x+1}$t=x+1​ back:

$$\log\left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right| + C$$log​x+1​+2x+1​−2​​+C