How do you integrate $\int \frac{1}{\sqrt{1 + x}}$ $int (1) / (sqrt(1 + x))$ ?

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Answer 1 · 2018-05-28T14:27:37

$\int \frac{1}{\sqrt{x + 1}} d x = 2 \sqrt{x + 1} + c$ $int1/sqrt(x+1)dx=2sqrt(x+1)+c$

$int1/sqrt(x+1)dx=2int((x+1)')/(2sqrt(x+1))dx=$

$2int(sqrt(x+1))'dx=2sqrt(x+1)+c$ $color(white)(aa)$ , $c$ $in$ $RR$

Answer 2 · 2018-05-28T14:28:15

$2sqrt(1+x)+C$

This function is very close to $sqrt(\frac{1}{x})$ , whose integral is $2sqrt(x)$ . In fact,

$\frac{d}{dx} 2sqrt(x) = 2\frac{d}{dx} sqrt(x) =2\frac{1}{2sqrt(x)} = \frac{1}{sqrt(x)}$

In our integral, you can substitute $t=x+1$ , which implies $dt=dx$ , since this is only a translation. So, you'd have

$\int \frac{1}{sqrt(t)} dt = 2sqrt(t)+C = 2sqrt(1+x) + C$

How do you integrate int (1) / (sqrt(1 + x))∫1√1+xint (1) / (sqrt(1 + x))?