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

The integral of 3/(sqrt(x+1)) is 6sqrt(x+1)+C

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

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

6 x + 1 + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{x + 1} d x

Apply u-substitution

: \int 2 d u

= 3 \cdot \int 2 d u

Integral of a constant:

\int a d x = a x

= 3 \cdot 2 u

Substitute back

u = x + 1

= 3 \cdot 2 x + 1

Simplify

= 6 x + 1

Add a constant to the solution

= 6 x + 1 + C

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

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

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

\int 8 x^{5} - 5 x^{4} - 4 x^{3} - 6 x^{2} - 2 x - 3 d x

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

What is the integral of 3/(sqrt(x+1)) ?
The integral of 3/(sqrt(x+1)) is 6sqrt(x+1)+C