Int 1/1+x^6 =?

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Int 1/1+x^6 =?

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Answer 1 · 2018-04-03T15:24:04

$\frac{7 x + x^{7}}{7}$ $(7x+x^7)/7$

Explanation:

$\int_{0}^{1} \frac{1}{1} + x^{6} = \int_{0}^{1} 1 + x^{6}$ $int_0^1 1/1+x^6=int_0^1 1+x^6$
${[1 . x]}_{0}^{1} + {[\frac{x^{6 + 1}}{6 + 1}]}_{0}^{1} = {[x]}_{0}^{1} + \frac{1}{7} {[x^{7}]}_{0}^{7}$ $[1.x]_0^1+[(x^(6+1))/(6+1)]_0^1=[x]_0^1+1/7[x^7]_0^1$
$x + \frac{1}{7} x^{7} = \frac{7 x + x^{7}}{7}$ $x+1/7x^7=(7x+x^7)/7$

Answer 2 · 2018-04-03T15:25:45

You need to specify if it was

$\int \frac{1}{1 + x^{6}} d x$ $int 1/(1+x^6)dx$

or

$\int \frac{1}{{(1 + x)}^{6}} d x$ $int 1/(1+x)^6dx$

Because in the 2nd case it's easier and is actually solveable whereas the 1st one is harder.
So the solution to $\int \frac{1}{{(1 + x)}^{6}} d x$ $int 1/(1+x)^6dx$ is:

we will take u as our new variable such as $u = (1 + x)$ $u=(1+x)$ and $d u = d x$ $du=dx$

which then gives us $\int \frac{1}{u^{6}} d u$ $int 1/u^6du$
and the answer is $- \frac{1}{5 u^{5}} +$ $-1/(5u^5) +$ constant

so the answer is $- \frac{1}{5 {(1 + x)}^{5}} +$ $-1/(5(1+x)^5) +$ constant

for solving the first case you need to consult better sites such as http://www.wolframalpha.com

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Int 1/1+x^6 =?

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