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Indefinite Integration

Section: Mathematics

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Question : 1 of 89

Marks: +1, -0

If

\int x^{49} \left[ \tan^{-1} x^{50} + \frac{x^{50}}{1+x^{100}} \right] dx = \frac{x^n}{k} f(x) + c

f(x) - f(\sqrt[k]{x^n}) =

[AP EAPCET 24 May 2025 S1]

$k + n$
$k - n$
$\frac{1}{k}$
$\frac{1}{n}$

Solution:

Given,

I = \int x^{49} \left[ \tan^{-1}(x^{50}) + \frac{x^{50}}{1+x^{100}} \right] dx

= \frac{x^n}{k} f(x) + c

Let

u = x^{50}

, then

du = 50 x^{49} dx

\text{so}, I = \int x^{49} \left[ \tan^{-1} x^{50} + \frac{x^{50}}{1+x^{100}} \right] dx

= \int x^{49} \left[ \tan^{-1}(u) + \frac{u}{1+u^2} \right] \cdot \frac{du}{50 x^{49}}

= \frac{1}{50} \int \left( \tan^{-1}(u) + \frac{u}{1+u^2} \right) du

= \frac{1}{50} \left[ \int \tan^{-1}(u) du + \int \frac{u}{1+u^2} du \right]

= \frac{1}{50} \left[ u \tan^{-1}(u) - \int \frac{u}{1+u^2} du + \int \frac{u}{1+u^2} du \right]

[using product rule]

= \frac{1}{50} \left[ u \tan^{-1}(u) \right] + c

= \frac{1}{50} \cdot x^{50} \cdot \tan^{-1}(x^{50}) + c

Comparing to original equation, we get

n = 50, k = 50, f(x) = \tan^{-1}(x^{50})

Now,

f(x) - f(\sqrt[k]{x^n})

= f(x) - f\left(x^{\frac{n}{k}}\right)

= \tan^{-1}(x^{50}) - f\left(x^{\frac{50}{50}}\right)

= \tan^{-1}(x^{50}) - f(x)

= \tan^{-1}(x^{50}) - \tan^{-1}(x^{50}) = 0

\therefore f(x) - f(\sqrt[k]{x^n}) = 0

Now,

k - n = 50 - 50 = 0

From Eqs. (i) and (ii), we get

f(x) - f(\sqrt[k]{x^n}) = k - n

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