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Integration

I'm unfamiliar on how to tackle these questions, so any help would be appreciated.

Integrate:

1. (1+x^2)^0.5

2. 1/(x^2+x-6)^1.5
Reply 1
1. I=1+x2  dx=(x)1+x2  dx=x1+x2x(1+x2)  dx=x1+x2x21+x2  dx=x1+x2(1+x2)11+x2  dx=x1+x21+x2  dx+11+x2  dx=x1+x2I+J.\begin{aligned}1. ~ I & = \int\sqrt{1+x^2}\;{dx} \\& = \int(x)'\sqrt{1+x^2}\;{dx} \\& = x\sqrt{1+x^2}-\int x\left(\sqrt{1+x^2}\right)'\;{dx} \\& = x\sqrt{1+x^2}-\int\frac{x^2}{\sqrt{1+x^2}}\;{dx} \\& = x\sqrt{1+x^2}-\int\frac{(1+x^2)-1}{\sqrt{1+x^2}}\;{dx} \\& = x\sqrt{1+x^2}-\int \sqrt{1+x^2}\;{dx}+\int\frac{1}{\sqrt{1+x^2}}\;{dx} \\& = x\sqrt{1+x^2}-I+J.\end{aligned}

Rearranging and recognizing that J is the definition of the inverse hyperbolic sine, we have:

2I=x1+x2+sinh1x+k2I = x\sqrt{1+x^2}+\sinh^{-1}{x}+k so I=12x1+x2+12sinh1x+kI = \frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\sinh^{-1}{x}+k.
You can do the second one in the exact same way, by completing the square first.
(edited 12 years ago)
Original post by Piecewise
1. I=1+x2  dx=(x)1+x2  dx=x1+x2x(1+x2)  dx=x1+x2x21+x2  dx=x1+x2(1+x2)11+x2  dx=x1+x21+x2  dx+11+x2  dx=x1+x2I+J.\begin{aligned}1. ~ I & = \int\sqrt{1+x^2}\;{dx} \\& = \int(x)'\sqrt{1+x^2}\;{dx} \\& = x\sqrt{1+x^2}-\int x\left(\sqrt{1+x^2}\right)'\;{dx} \\& = x\sqrt{1+x^2}-\int\frac{x^2}{\sqrt{1+x^2}}\;{dx} \\& = x\sqrt{1+x^2}-\int\frac{(1+x^2)-1}{\sqrt{1+x^2}}\;{dx} \\& = x\sqrt{1+x^2}-\int \sqrt{1+x^2}\;{dx}+\int\frac{1}{\sqrt{1+x^2}}\;{dx} \\& = x\sqrt{1+x^2}-I+J.\end{aligned}

Rearranging and recognizing that J is the definition of the inverse hyperbolic sine, we have:

2I=x1+x2+sinh1x+k2I = x\sqrt{1+x^2}+\sinh^{-1}{x}+k so I=12x1+x2+12sinh1x+kI = \frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\sinh^{-1}{x}+k.
You can do the second one in the exact same way, by completing the square first.


nice method! (just one teeny edit :smile: )

+ repped
Do these types of integrals fall withing the purview of the Edexcel C4 specification ?
Original post by Ari Ben Canaan
Do these types of integrals fall withing the purview of the Edexcel C4 specification ?


nope.

You are not expected to know hyperbolic functions for A level
Integration; the intermixing of people or groups previously segregated.

hope that helps, maths was never one of my strong points.
Reply 6
Either as above or make a trig substitution
Reply 7
Original post by Plato's Trousers
nice method! (just one teeny edit :smile: )

+ repped
Thank you! :)

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