calculus integral question
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Have you tried trigonometric substitution?
Please show us what you have tried and exactly where you are stuck.
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HallsofIvy
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If you are taking a Calculus class then you should recognize \(\displaystyle \sqrt{u^2+ 1}\) as one of the basic integrals (perhaps given in a "table of integrals"). As Subhotosh Kahn suggested, try a trig substitution.
Surely you remember "\(\displaystyle sin^2(\theta)+ cos^2(\theta)= 1\)"? Divide both sides by \(\displaystyle cos^2(\theta)\) to get a similar trig identity \(\displaystyle \frac{sin^2(\theta)}{cos^2(\theta)}+ 1= \frac{1}{cos^2(\theta)}\) or \(\displaystyle tan^2(\theta)+ 1= sec^2(\theta)\). Compare that to your \(\displaystyle 7x^2+ 4= 4\left(\frac{7}{4}x^2+ 1\right)\). Does that suggest letting \(\displaystyle \frac{\sqrt{7}}{2}x= tan(\theta)\)?
Surely you remember "\(\displaystyle sin^2(\theta)+ cos^2(\theta)= 1\)"? Divide both sides by \(\displaystyle cos^2(\theta)\) to get a similar trig identity \(\displaystyle \frac{sin^2(\theta)}{cos^2(\theta)}+ 1= \frac{1}{cos^2(\theta)}\) or \(\displaystyle tan^2(\theta)+ 1= sec^2(\theta)\). Compare that to your \(\displaystyle 7x^2+ 4= 4\left(\frac{7}{4}x^2+ 1\right)\). Does that suggest letting \(\displaystyle \frac{\sqrt{7}}{2}x= tan(\theta)\)?