The integral dx/(x^2(x^4+1)^(3/4)) equals __. Option 1: ((x^4+1)/x^4)^(1/4) + C Option 2: (x^4+1)^(1/4) + C Option 3: -(x^4+1)^(1/4) + C Option 4: -((x^4+1)/x^4)^(1/4) + C

Answer: -((x^4+1)/x^4)^(1/4) + C. Solution: Rewrite denominator: x^2(x^4+1)^(3/4) = x^5(1 + 1/x^4)^(3/4). Let u = 1 + 1/x^4, du = -4/x^5dx. Integral = -1/4 u^(-3/4)du = -u^(1/4) + C = -((x^4+1)/x^4)^(1/4) + C.

Calculus

Integrals

Hard

core

The integral $\int \frac{dx}{x^2(x^4+1)^{\frac{3}{4}}}$ equals __________.

$\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}} + C$

$(x^4+1)^{\frac{1}{4}} + C$

$-(x^4+1)^{\frac{1}{4}} + C$

$-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}} + C$

✅ Correct Option: 4

Rewrite denominator: $x^2(x^4+1)^{3/4} = x^5\left(1 + \frac{1}{x^4}\right)^{3/4}$ .

Let $u = 1 + \frac{1}{x^4}$ , $du = -\frac{4}{x^5}dx$ .

Integral $= -\frac{1}{4}\int u^{-3/4}du = -u^{1/4} + C = -\left(\frac{x^4+1}{x^4}\right)^{1/4} + C$ .

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CUET Mathematics Past Year Questions (Topic-Wise)

The integral $\int \frac{dx}{x^2(x^4+1)^{\frac{3}{4}}}$ equals __________.