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CBSE Sample Paper 2024-25 · Section A · 1 mark

Q12: ∫dxx3(1+x4)1/2\int\frac{dx}{x^3(1 + x^4)^{1/2}}

Sample 2024-25Ch 7 · Integrals
The question

∫dxx3(1+x4)12\displaystyle\int\frac{dx}{x^3(1 + x^4)^{\frac{1}{2}}} equals

(A) −12x21+x4+c-\frac{1}{2x^2}\sqrt{1 + x^4} + c

(B) 12x1+x4+c\frac{1}{2x}\sqrt{1 + x^4} + c

(C) −14x1+x4+c-\frac{1}{4x}\sqrt{1 + x^4} + c

(D) 14x21+x4+c\frac{1}{4x^2}\sqrt{1 + x^4} + c

Exam answer

Let I=∫dxx3(1+x4)12I = \displaystyle\int\frac{dx}{x^3(1 + x^4)^{\frac{1}{2}}}

I=∫dxx5(1+1x4)12I = \int\frac{dx}{x^5\left(1 + \frac{1}{x^4}\right)^{\frac{1}{2}}}
Why?

Take x4x^4 out of the bracket: 1+x4=x4(1+1x4)1 + x^4 = x^4\left(1 + \frac{1}{x^4}\right)⁠, and (x4)12=x2(x^4)^{\frac12} = x^2⁠. So

x3(1+x4)12=x3(x4(1+1x4))12=x3⋅(x4)12(1+1x4)12=x3⋅x2(1+1x4)12=x5(1+1x4)12\begin{aligned} &x^3(1 + x^4)^{\frac12} \\ &= x^3\left(x^4\left(1 + \frac{1}{x^4}\right)\right)^{\frac12} \\ &= x^3 \cdot (x^4)^{\frac12}\left(1 + \frac{1}{x^4}\right)^{\frac12} \\ &= x^3 \cdot x^2\left(1 + \frac{1}{x^4}\right)^{\frac12} \\ &= x^5\left(1 + \frac{1}{x^4}\right)^{\frac12} \end{aligned}

(x3⋅x2=x3+2=x5x^3 \cdot x^2 = x^{3 + 2} = x^5⁠.)

This brings out 1x5\frac{1}{x^5}⁠, which is (up to a constant) the derivative of 1x4\frac{1}{x^4}⁠.

Put 1+1x4=t1 + \frac{1}{x^4} = t⁠. Then −4x5 dx=dt-\frac{4}{x^5}\,dx = dt⁠.

Why?

Write 1x4=x−4\frac{1}{x^4} = x^{-4}⁠. Its derivative is −4x−5-4x^{-5}⁠, so

t=1+x−4dtdx=0+(−4)x−4−1=−4x−5=−4x5dt=−4x5 dx−14 dt=dxx5(÷ (−4))\begin{aligned} t &= 1 + x^{-4} \\ \frac{dt}{dx} &= 0 + (-4)x^{-4 - 1} \\ &= -4x^{-5} = -\frac{4}{x^5} \\[4pt] dt &= -\frac{4}{x^5}\,dx \\ -\frac14\,dt &= \frac{dx}{x^5} \quad (\div\,(-4)) \end{aligned} dxx5=−14 dt\frac{dx}{x^5} = -\frac{1}{4}\,dt

So in II⁠, replace dxx5\frac{dx}{x^5} by −14 dt-\frac14\,dt and (1+1x4)12\left(1 + \frac{1}{x^4}\right)^{\frac12} by t12t^{\frac12}⁠:

I=∫1t12⋅(−14)dt=−14∫t−12 dt\begin{aligned} I &= \int \frac{1}{t^{\frac12}} \cdot \left(-\frac14\right)dt \\ &= -\frac14\int t^{-\frac12}\,dt \end{aligned}
I=−14∫t−12 dt=−14×2t+c=−121+1x4+c=−12x4+1x4+c=−12x21+x4+c\begin{aligned} I &= -\frac{1}{4}\int t^{-\frac{1}{2}}\,dt \\ &= -\frac{1}{4} \times 2\sqrt t + c \\ &= -\frac{1}{2}\sqrt{1 + \frac{1}{x^4}} + c \\ &= -\frac{1}{2}\sqrt{\frac{x^4 + 1}{x^4}} + c \\ &= -\frac{1}{2x^2}\sqrt{1 + x^4} + c \end{aligned}
Why?

Power rule with n=−12n = -\frac12⁠:

∫t−12 dt=t1212=2t\int t^{-\frac12}\,dt = \frac{t^{\frac12}}{\frac12} = 2\sqrt t

Then put tt back and use x4=x2\sqrt{x^4} = x^2 in the denominator.

−14×2=−24=−121+1x4=x4x4+1x4=x4+1x4x4+1x4=x4+1x4=1+x4x2−12⋅1+x4x2=−12x21+x4\begin{aligned} -\frac14 \times 2 &= -\frac24 = -\frac12 \\[4pt] 1 + \frac{1}{x^4} &= \frac{x^4}{x^4} + \frac{1}{x^4} \\ &= \frac{x^4 + 1}{x^4} \\[4pt] \sqrt{\frac{x^4 + 1}{x^4}} &= \frac{\sqrt{x^4 + 1}}{\sqrt{x^4}} \\ &= \frac{\sqrt{1 + x^4}}{x^2} \\[4pt] &-\frac12 \cdot \frac{\sqrt{1 + x^4}}{x^2} \\ &= -\frac{1}{2x^2}\sqrt{1 + x^4} \end{aligned}

Watch out: the factor outside is 12x2\frac{1}{2x^2}⁠, with x2x^2⁠, not xx as in options (B) and (C).

Answer: (A) −12x21+x4+c-\dfrac{1}{2x^2}\sqrt{1 + x^4} + c

Where this comes from

Chapter 7, Integrals: Integration by substitution.

Question type: Substitution

How to spot it: click to see

Type 2. Substitution: when the derivative is hiding in the question

Spot it: a function and its derivative both appear, like log⁡x\log x and 1x\frac{1}{x}⁠, or x2+1x^2 + 1 and xx⁠.

Here: Taking x4x^4 out of the root leaves 1x5\frac{1}{x^5} beside (1+1x4)1/2\left(1 + \frac{1}{x^4}\right)^{1/2}⁠, and 1x5\frac{1}{x^5} is the derivative of 1x4\frac{1}{x^4} apart from the −4-4⁠. So put 1+1x4=t1 + \frac{1}{x^4} = t⁠, and the integral is −14∫t−1/2 dt-\frac14\int t^{-1/2}\,dt⁠.

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