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Integration

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Re: Integration

Dear mathstudent200, Please remove your article ”integration” – your question is no encyclopedia entry! The integrand is a linear polynomial divided by the square root of another linear polynomial. You could execute the division by first writing the dividend in the form (1/2)[(4x-1)-5]124x15(1/2)[(4x-1)-5]; divide separately the minuend and the subtrahend. Then it’s easy to integrate separately both quotients! Greetings, Jussi

Re: Integration

Usually, if you have an integrand that contains a problematic nested function, like the polynomial inside the square root, we can usually simplify it greatly by letting a single variable u represent the value of that function, and rewriting the whole integral in terms of uuu. In order to integrate the new expression written in terms of uuu, we also need du/dxdudxdu/dx.

In your case, the problematic function is the square root of a polynomial in xxx, so we let u=4x-1u4x1u=4x-1 in order to turn the problematic square root into a simple power of a single variable u1/2superscriptu12u^{{1/2}}. We need to write 2x-32x32x-3 in terms of uuu as well. We can solve for xxx in terms of uuu by using our equation u=4x-1u4x1u=4x-1. This gives us x=(u+1)/4xu14x=(u+1)/4, so 2x-32x32x-3 becomes (1/2)u-5/212u52(1/2)u-5/2.

Dividing by u1/2superscriptu12u^{{1/2}}, our integrand is now (1/2)u1/2-(5/2)u-1/2dx12superscriptu1252superscriptu12dx(1/2)u^{{1/2}}-(5/2)u^{{-1/2}}dx. We cannot integrate this function of uuu over dxdxdx, so we must find out how dxdxdx is related to dududu. Differentiating u=4x-1u4x1u=4x-1 gives us du/dx=4dudx4du/dx=4, or dx=(1/4)dudx14dudx=(1/4)du. Replacing dxdxdx with (1/4)du14du(1/4)du in our integral allows us to integrate it, as it is just a sum of two power rules. Once we have our integrated function in terms of uuu, we can replace uuu with 4x-14x14x-1 and simplify to get our function in terms of xxx.

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