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$$\int \frac{dx}{\sqrt {7-6x}-x^2}$$

Have no idea how to start with this. How do i make a perfect square of the variable?

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    Check this: https://www.wolframalpha.com/input/?i=integral+1%2F(sqrt(7-6x)-x%5E2) Maybe the square root is to be longer. – 2017-02-13
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    @Rafa no if the sqrt was longer i wouldn't have posted the problem here. http://imgur.com/ttEF2gE 12. – 2017-02-13
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    Two of them seem pretty elementary. The fourth it's a headhache "as is" or in the way of the others if you extend the square root symbol.You may consider somebody typing from handwriting or by means an app to copy algebra. – 2017-02-13
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    How is this question off topic? – 2017-02-15

2 Answers 2

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I assume you are trying to solve the integral $$I:=\int \frac{\mathrm d x}{\sqrt{7-6x-x^2}}$$ because you mentioned to make a perfect square. Since $$7-6x-x^2= 16-(x+3)^2= \frac{1}{16}\left(1- \left( \frac{x+3}{4}\right)^2\right)$$ the integral becomes

$$I= \int \frac{\mathrm d x}{\frac{1}{4}\sqrt{1- \left( \frac{x+3}{4}\right)^2 }} = \arcsin \left( \frac{x+3}{4}\right).$$

In the last step you need to substitute $t= \frac{x+3}{2}$.

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    Lol, that would make more sense I suppose. You have my vote :) – 2017-02-13
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    @BrevanEllefsen Thank you. Its just a guess ;) – 2017-02-13
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    @Niklas No the sqrt isn't over the whole denominaor. http://imgur.com/ttEF2gE 12 – 2017-02-13
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    @mettledmike Okay, where does this problem come from? – 2017-02-13
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    @Niklas from my school textbook – 2017-02-15
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    Well, I think it is a typo then. Your original integral is to difficult (and nonelementar). – 2017-02-15
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One viable route is to make a substitution $$\int \frac{dx}{\sqrt {7-6x}-x^2}$$ $u=\sqrt{7-6x} \implies x=(7-u^2)/6 \implies dx = u/3\, du$ $$=\frac{1}{3}\int \frac{u}{u-\frac{(u^2-7)^2}{36}}\, du$$
Now simplify the denominator and run partial fraction decomposition. Might not be pretty, but it's definitely solvable since the quartic polynomial is solvable. Worst case scenario just let $a_i$ be the $i$th root and express in terms of $a_i$