# How do I begin integrating this one? $\int_0^x\frac{1}{1-t-t^2}dt$

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Anyways...How do I begin integrating this one? $$\int_0^x\frac{1}{1-t-t^2}dt$$

I don't think this is a u/du substitution. Do I change it to this? $$\int_0^x (1-t-t^2)^{-1} dt=$$

Disclaimer: I am not a student posting his homework assignment. I am an adult learning Calculus.

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Hint: Complete the square. – Amzoti Dec 6 '13 at 13:54
Substitute $u = \left(t+\frac12\right)$. You get $\int_{1/2}^{x+1/2} \frac{du}{5/4 - u^2}$, which you probably know how to handle. Or do a partial fraction decomposition. – Daniel Fischer Dec 6 '13 at 13:54
Look in your textbook for "partial fractions" .... there may be a section called "integrating rational functions" but maybe only stuck in the middle of "methods of integration". – GEdgar Dec 6 '13 at 14:29
Ok, so the bottom line is this is an advanced integration technique that I have yet to encounter? I will look it up, or revisit at a later date. Thanks! – JackOfAll Dec 6 '13 at 22:51
Not sure where to go from here: $$\int_0^x\frac{1}{\frac{5}{4}-u^2}du$$ Also not sure how to change limits of integration. What topic should I look up, in order how to do this? I have not encountered this method when learning basic u/du substitution for indefinite integrals – JackOfAll Dec 8 '13 at 17:31

Hint: Write the integrand as $\dfrac{1}{5/4- (t+1/2)^2}$
Depends on the value of $x$. – bwv869 Dec 6 '13 at 16:38
No, actually it is not advanced at all. I have just completed the square, now you just need to substitute $u=t + 1/2$ and split into partial fractions, then you can integrate. You will need to use $\int \frac{1}{x} \mathrm{d}x = \ln{x}+C$ – bwv869 Dec 6 '13 at 23:30
Not sure where to go from here: $$\int_0^x\frac{1}{\frac{5}{4}-u^2}du$$ Also not sure how to change limits of integration. What topic should I look up, in order how to do this? I have not encountered this method when learning basic u/du substitution for indefinite integrals – JackOfAll Dec 8 '13 at 17:29