What is the integral of (-1)/(sqrt(1-(4t+1)^2)) ?

The integral of (-1)/(sqrt(1-(4t+1)^2)) is -1/4 arcsin(4t+1)+C

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integral of (-1)/(sqrt(1-(4t+1)^2))

\int \frac{- 1}{1 - ( 4 t + 1 ) ^{2}} d t

- \frac{1}{4} arcsin (4 t + 1) + C

Solution steps

\int \frac{- 1}{1 - ( 4 t + 1 ) ^{2}} d t

Take the constant out:

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

= - \int \frac{1}{1 - ( 4 t + 1 ) ^{2}} d t

Apply u-substitution

: \int \frac{1}{4 1 - u ^{2}} d u

= - \int \frac{1}{4 1 - u ^{2}} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{1}{1 - u ^{2}} d u

Use the common integral:

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

= - \frac{1}{4} arcsin (u)

Substitute back

u = 4 t + 1

= - \frac{1}{4} arcsin (4 t + 1)

Add a constant to the solution

= - \frac{1}{4} arcsin (4 t + 1) + C

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What is the integral of (-1)/(sqrt(1-(4t+1)^2)) ?
The integral of (-1)/(sqrt(1-(4t+1)^2)) is -1/4 arcsin(4t+1)+C