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Worked solution

Expand after the specified linear substitution

Use the substitution y=1−2xy=1-2x to find ∫x2+11−2x dx\displaystyle\int\dfrac{x^2+1}{\sqrt{1-2x}}\,dx. marks

[2013/HCI/I/9(a)]

CJC · Review Problems · Question 4

Answer

−120(1−2x)5/2+16(1−2x)3/2−541−2x+C-\frac1{20}(1-2x)^{5/2}+\frac16(1-2x)^{3/2}-\frac54\sqrt{1-2x}+C.

Specified substitution

With y=1−2xy=1-2x, x=(1−y)/2x=(1-y)/2 and dx=−dy/2dx=-dy/2. The transformed integral is −∫5−2y+y28y dy-\int\dfrac{5-2y+y^2}{8\sqrt y}\,dy. Follow the source guidance to expand and simplify: integrate −58y−1/2+14y1/2−18y3/2-\frac58y^{-1/2}+\frac14y^{1/2}-\frac18y^{3/2}. The result is −54y1/2+16y3/2−120y5/2+C-\frac54y^{1/2}+\frac16y^{3/2}-\frac1{20}y^{5/2}+C. Thus the printed ordering is −120(1−2x)5/2+16(1−2x)3/2−54(1−2x)1/2+C-\frac1{20}(1-2x)^{5/2}+\frac16(1-2x)^{3/2}-\frac54(1-2x)^{1/2}+C.

How did this part go?

Common mistakes

The Jacobian dx=−dy/2 supplies both a minus sign and a factor 1/2; the real integrand requires x<1/2.

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