What is the integral of 1/(sqrt(x+3)-1) ?

The integral of 1/(sqrt(x+3)-1) is -ln|sqrt(x+3)+1|+ln|sqrt(x+3)-1|+2sqrt(x+3)+ln|x+2|+C

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integral of 1/(sqrt(x+3)-1)

\int \frac{1}{x + 3 - 1} d x

- ln x + 3 + 1 + ln x + 3 - 1 + 2 x + 3 + ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{1}{x + 3 - 1} d x

\frac{1}{x + 3 - 1} = \frac{x + 3 + 1}{( x + 3 ) ^{2} - ( 1 ) ^{2}}

= \int \frac{x + 3 + 1}{( x + 3 ) ^{2} - 1 ^{2}} d x

Expand

\frac{x + 3 + 1}{( x + 3 ) ^{2} - 1 ^{2}} : \frac{x + 3}{( x + 3 ) ^{2} - 1 ^{2}} + \frac{1}{( x + 3 ) ^{2} - 1 ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x + 3}{( x + 3 ) ^{2} - 1 ^{2}} d x + \int \frac{1}{( x + 3 ) ^{2} - 1 ^{2}} d x

\int \frac{x + 3}{( x + 3 ) ^{2} - 1 ^{2}} d x = - ln x + 3 + 1 + ln x + 3 - 1 + 2 x + 3

\int \frac{1}{( x + 3 ) ^{2} - 1 ^{2}} d x = ln ∣ x + 2 ∣

= - ln x + 3 + 1 + ln x + 3 - 1 + 2 x + 3 + ln ∣ x + 2 ∣

Add a constant to the solution

= - ln x + 3 + 1 + ln x + 3 - 1 + 2 x + 3 + ln ∣ x + 2 ∣ + C

\int sin (x) + C d x

\int e^{8 + x} d x

\int_{1}^{3} (2 x^{2} - 3 x + 5) d x

e^{13 t} sin (t)

x \frac{d y}{d x} + y = 6 x + 1, y (1) = 9

What is the integral of 1/(sqrt(x+3)-1) ?
The integral of 1/(sqrt(x+3)-1) is -ln|sqrt(x+3)+1|+ln|sqrt(x+3)-1|+2sqrt(x+3)+ln|x+2|+C