## $\int \frac{(x + x^{2/3} + x^{1/6})}{x(1 + x^{1/3})} dx$ is equal to

Q: $\large \int \frac{(x + x^{2/3} + x^{1/6})}{x(1 + x^{1/3})} dx$ is equal to

(A) $\large \frac{3}{2}x^{2/3} + 6 tan^{-1}(x^{1/6}) + c$

(B) $\large \frac{3}{2}x^{2/3} – 6 tan^{-1}(x^{1/6}) + c$

(C) $\large \frac{3}{2}x^{2/3} + tan^{-1}(x^{1/6}) + c$

(D) none of these

Sol:  Substituting x = z6, dx = 6 z5dz , we have

$\large I = \int \frac{6 z^5 (z^6 + z^4 + z)}{z^6 (1+z^2)}dz$

$\large = \int \frac{6 (z^5 + z^3 + 1)}{(1+z^2)}dz$

$\large = \int 6 z^3 dz + \int (\frac{6 dz}{z^2 + 1})$

$\large = 6\frac{z^4}{4} + 6 tan^{-1}z + c$

$\large \frac{3}{2}x^{2/3} + 6 tan^{-1}(x^{1/6}) + c$

Hence (A) is the correct answer.

## $\int \frac{x^2 -1}{(x^2 +1)\sqrt{x^4 +1}} dx$ is equal to

Q: $\large \int \frac{x^2 -1}{(x^2 +1)\sqrt{x^4 +1}} dx$ is equal to

(A) $\large sec^{-1}(\frac{x^2 + 1}{\sqrt{2}x}) + c$

(B) $\large \frac{1}{\sqrt{2}} sec^{-1}(\frac{x^2 + 1}{\sqrt{2}x}) + c$

(C) $\large \frac{1}{\sqrt{2}} sec^{-1}(\frac{x^2 + 1}{\sqrt{2}}) + c$

(D) none of these

Sol: $\large I = \int \frac{x^2(1 -\frac{1}{x^2})}{x^2(x +\frac{1}{x})\sqrt{x^2 + \frac{1}{x^2}}} dx$

Let $\large x + \frac{1}{x} = z$ ; $\large (1-\frac{1}{x^2})dx = dz$

$\large I = \int \frac{dz}{z \sqrt{z^2 -2}}$

$\large = \frac{1}{\sqrt{2}} sec^{-1}\frac{z}{\sqrt{2}}$

$\large = \frac{1}{\sqrt{2}} sec^{-1}(\frac{x^2 + 1}{\sqrt{2}x}) + c$

## If f'(x) = x |sinx| ∀ x  ∈ (0, π) and f(0) = 1/√3 , then f(x) will be

Q: If f'(x) = x |sinx| ∀ x  ∈ (0, π) and f(0) = 1/√3 , then f(x) will be

(A) – x cosx + sinx + 1/√3

(B) 1/√3 – x cosx + sinx

(C) sinx + x cosx – 2/√3

(D) sinx + x cosx + 2/√3

Sol. f'(x) = x |sinx| ∀ x  ∈ (0, π)

$\large \int f'(x) dx = \int x sinx dx$

$\large f(x) = – x cosx + \int cosx dx + c$

$\large = – x cosx + sinx + c$

f(0) = c = 1/√3

$\large f(x) = – x cosx + sinx + \frac{1}{\sqrt{3}}$

Hence (A) is the correct answer.

## $\int \frac{cosx (1 + 4 cos2x)}{sinx + 4 sinx cos^2 x} dx$ is equal to

Q: $\large \int \frac{cosx (1 + 4 cos2x)}{sinx + 4 sinx cos^2 x} dx$ is equal to

(A) $\large \frac{1}{2} ln|sin^2 x + sin^2 2x| + c$

(B) $\large \frac{1}{2}ln |sinx| + \frac{1}{2} ln|sin x + 4sinx cos^2x| + c$

(C) $\large \frac{1}{2} ln|sin^2 x – sin^2 2x| + c$

(D) $\large \frac{1}{2}ln |sinx| – ln|sin x + 4sinx cos^2x| + c$

Sol: $\large \int \frac{cosx (1 + 4 cos2x)}{sinx + 4 sinx cos^2 x} dx$

Put cosx = t

$\large = – \int \frac{t (1 + 4(2t^2 – 1))}{1-t^2 + 4 (1-t^2)t^2} dt$

$\large = \int \frac{t(8t^2 – 3)}{4t^4 – 3t^2 – 1} dt$

$\large = \frac{1}{2} ln |4t^4 – 3t^2 – 1| + c$

$\large = \frac{1}{2}ln |sinx| + \frac{1}{2} ln|sin x + 4sinx cos^2x| + c$

## $\int ( x + \frac{1}{x})^{n+5} (\frac{x^2 -1}{x^2})dx$ is equal to

Q: $\int ( x + \frac{1}{x})^{n+5} (\frac{x^2 -1}{x^2})dx$ is equal to

(A) $\frac{(x+ \frac{1}{x})^{n+6}}{n+6} + c$

(B) $(\frac{x^2 + 1}{x^2})^{n+6} (n+6) + c$

(C) $(\frac{x}{x^2 + 1})^{n+6} (n+6) + c$

(D) none of these

Sol: Let $(x+ \frac{1}{x}) = z$ then $(1 – \frac{1}{x^2})dx = dz$

$I = \int z^{n+5} dz = \frac{z^{n+6}}{n+6} + c$

$= \frac{(x + \frac{1}{x})^{n+6}}{n+6} + c$

Hence (A) is the correct answer.