(x + a/x)/(x^2 + a^2/x^2) dx, a > 0 is equal to (where C is an arbitrary constant) Option 1: 1/4_e, x^4 + a^2, + 1/2^(-1)x^2/a + C Option 2: 1/2_e, x^2 + a, + 1/4^(-1)x^4/a^2 + C Option 3: 1/4_e, x^4 + a^2, + a/2^(-1)x^2/a + C Option 4: 1/2_e, x^2 + a, + a/4^(-1)x^4/a^2 + C

Answer: 1/4_e, x^4 + a^2, + 1/2^(-1)x^2/a + C. Solution: Multiply numerator and denominator by x^2 to get (x^3 + ax)/(x^4 + a^2)dx. The first part gives 1/4, x^4+a^2, . For the second, put u=x^2: a du/(2(u^2+a^2)) = 1/2^(-1)(x^2/a). Adding gives option 1.

Calculus

Integrals

Hard

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$\displaystyle\int \dfrac{x + \dfrac{a}{x}}{x^2 + \dfrac{a^2}{x^2}}\,dx, a > 0$ is equal to (where C is an arbitrary constant)

$\dfrac{1}{4}\log_e|x^4 + a^2| + \dfrac{1}{2}\tan^{-1}\dfrac{x^2}{a} + C$

$\dfrac{1}{2}\log_e|x^2 + a| + \dfrac{1}{4}\tan^{-1}\dfrac{x^4}{a^2} + C$

$\dfrac{1}{4}\log_e|x^4 + a^2| + \dfrac{a}{2}\tan^{-1}\dfrac{x^2}{a} + C$

$\dfrac{1}{2}\log_e|x^2 + a| + \dfrac{a}{4}\tan^{-1}\dfrac{x^4}{a^2} + C$

✅ Correct Option: 1

Multiply numerator and denominator by $x^2$ to get $\int \dfrac{x^3 + ax}{x^4 + a^2}dx$ .

The first part gives $\dfrac{1}{4}\log|x^4+a^2|$ . For the second, put $u=x^2$ : $a\int \dfrac{du}{2(u^2+a^2)} = \dfrac{1}{2}\tan^{-1}(x^2/a)$ .

Adding gives option 1.

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Value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \log(\tan x)dx$ is

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Integral $I = \displaystyle\int \dfrac{e^{3\log x} - 3x^4}{e^{2\log x} + x^2}\, dx$ (where $x>0$ ) is equal to

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Match List-I with List-II

List-I	List-II
Indefinite integral	Solution (where 'c' is an arbitrary constant)
(A) $\displaystyle\int \sqrt{x^2-a^2} \, dx$	(I) $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + c$
(B) $\displaystyle\int \dfrac{1}{x^2-a^2} \, dx$	(II) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left[x+\sqrt{x^2+a^2}\right] + c$
(C) $\displaystyle\int \dfrac{1}{x^2+a^2} \, dx$	(III) $\dfrac{x}{2}\sqrt{x^2-a^2} - \dfrac{a^2}{2}\log\left[x+\sqrt{x^2-a^2}\right] + c$
(D) $\displaystyle\int \sqrt{x^2+a^2} \, dx$	(IV) $\dfrac{1}{2a}\log\left\vert \dfrac{x-a}{x+a} \right\vert + C$

Choose the correct answer from the options given below:

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The value of $\int_0^1 x e^x dx$ is:

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$\int \frac{\log_e x}{(1 + \log_e x)^2} dx$ is equal to

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The value of $\int_{-1}^{1} \dfrac{x^6}{6^x+1}\,dx$ is

∫x+axx2+a2x2 dx,a>0\displaystyle\int \dfrac{x + \dfrac{a}{x}}{x^2 + \dfrac{a^2}{x^2}}\,dx, a > 0∫x2+x2a2​x+xa​​dx,a>0 is equal to (where C is an arbitrary constant)

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$\displaystyle\int \dfrac{x + \dfrac{a}{x}}{x^2 + \dfrac{a^2}{x^2}}\,dx, a > 0$ is equal to (where C is an arbitrary constant)