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