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

The integral of 1/(sqrt(1-(x+1)^2)) is arcsin(x+1)+C

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

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

arcsin (x + 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{1 - u ^{2}} d u

= \int \frac{1}{1 - u ^{2}} d u

Use the common integral:

\int \frac{1}{1 - u ^{2}} d u = arcsin (u)

= arcsin (u)

Substitute back

u = x + 1

= arcsin (x + 1)

Add a constant to the solution

= arcsin (x + 1) + C

\frac{\partial}{\partial y} (ln (x^{2} + y^{2} + z^{2}))

\int \frac{( y - 1 ) ( y - 5 )}{y ^{3} - y ^{2} + 3 y + 5} d y

\int 2 + 6 x d x

n = 1 \sum \infty x^{n + 1}

\frac{\partial}{\partial x} (x \cdot e^{x \cdot y})

What is the integral of 1/(sqrt(1-(x+1)^2)) ?
The integral of 1/(sqrt(1-(x+1)^2)) is arcsin(x+1)+C