The value of _(-1)^1 x^6/(6^x+1) dx is Option 1: 3 Option 2: 6 Option 3: 2/7 Option 4: 1/7

Answer: 1/7. Solution: Let I = _(-1)^1x^6/(6^x+1)dx. Replacing x by -x gives I = _(-1)^1(6^x x^6)/(6^x+1)dx. Adding, 2I = _(-1)^1x^6 dx = 2/7. Hence I = 1/7.

Calculus

Integrals

Medium

core

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

$\dfrac{2}{7}$

$\dfrac{1}{7}$

✅ Correct Option: 4

Let $I = \int_{-1}^{1}\dfrac{x^6}{6^x+1}dx$ . Replacing $x$ by $-x$ gives $I = \int_{-1}^{1}\dfrac{6^x x^6}{6^x+1}dx$ .

Adding, $2I = \int_{-1}^{1}x^6\,dx = \dfrac{2}{7}$ .

Hence $I = \dfrac{1}{7}$ .

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If C is an arbitrary constant, then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\displaystyle\int a^x \, dx$	I.	$\sin^{-1}\dfrac{x}{a} + C$
B.	$\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}}$	II.	$\log_e \vert x + \sqrt{x^2 - a^2} \vert + C$
C.	$\displaystyle\int \dfrac{dx}{\sqrt{a^2 + x^2}}$	III.	$\dfrac{a^x}{\log_e a} + C$
D.	$\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}}$	IV.	$\log_e \vert x + \sqrt{a^2 + x^2} \vert + C$

Choose the correct answer from the options given below:

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Value of $\int (e^{2\log x} + e^{x\log 2} + e^{2\log 2})dx$ is

The value of ∫−11x66x+1 dx\int_{-1}^{1} \dfrac{x^6}{6^x+1}\,dx∫−11​6x+1x6​dx is

Solution

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