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Question
Prove that:
`int sqrt(x^2 - a^2)dx = x/2sqrt(x^2 - a^2) - a^2/2log|x + sqrt(x^2 - a^2)| + c`
Solution
Let I = `int sqrt(x^2 - a^2)dx`
I = `int sqrt(x^2 - a^2)*1dx`
I = `sqrt(x^2 - a^2)*int1dx - int[d/dx(sqrt(x^2 - a^2))*int1dx]dx`
I = `sqrt(x^2 - a^2)*x - int[1/(2sqrt(x^2 - a^2))*d/dx(x^2 - a^2)*x]dx`
I = `sqrt(x^2 - a^2)*x - int1/(2sqrt(x^2 - a^2))(2x - 0)*x dx`
I = `sqrt(x^2 - a^2)*x - intx/sqrt(x^2 - a^2)*x dx`
I = `xsqrt(x^2 - a^2) - int(x^2 - a^2 + a^2)/(sqrt(x^2 - a^2))dx`
I = `xsqrt(x^2 - a^2) - intsqrt(x^2 - a^2) dx - a^2intdx/(sqrt(x^2 - a^2)`
I = `xsqrt(x^2 - a^2) - I - a^2log|x + sqrt(x^2 - a^2)| + c_1`
∴ 2I = `xsqrt(x^2 - a^2) - a^2log|x + sqrt(x^2 - a^2)| + c_1`
∴ I = `x/2sqrt(x^2-a^2) - a^2/2log|x + sqrt(x^2 - a^2)| + c_1/2`
∴ `intsqrt(x^2 - a^2) dx = x/2sqrt(x^2 - a^2) - a^2/2log|x + sqrt(x^2 - a^2)| + c, "where" c = c_1/2`
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