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

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

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integral of 1/(sqrt(1-2x)+3)

\int \frac{1}{1 - 2 x + 3} d x

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

Solution steps

\int \frac{1}{1 - 2 x + 3} d x

Apply u-substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int \frac{1}{u + 3} d u

Apply u-substitution

= - \frac{1}{2} \cdot \int \frac{2 ( v - 3 )}{v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot 2 \cdot \int \frac{v - 3}{v} d v

Expand

\frac{v - 3}{v} : 1 - \frac{3}{v}

= - \frac{1}{2} \cdot 2 \cdot \int 1 - \frac{3}{v} d v

Apply the Sum Rule:

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

= - \frac{1}{2} \cdot 2 (\int 1 d v - \int \frac{3}{v} d v)

\int 1 d v = v

\int \frac{3}{v} d v = 3 ln ∣ v ∣

= - \frac{1}{2} \cdot 2 (v - 3 ln ∣ v ∣)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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