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

The integral of 1/(sqrt(1+(y-1)^2)) is ln|sqrt(y^2-2y+2)+y-1|+C

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

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

ln y^{2} - 2 y + 2 + y - 1 + C

Solution steps

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

Expand

1 + (y - 1)^{2} : y^{2} - 2 y + 2

= \int \frac{1}{y ^{2} - 2 y + 2} d y

Complete the square

y^{2} - 2 y + 2 : (y - 1)^{2} + 1

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = ln u^{2} + 1 + u

= ln u^{2} + 1 + u

Substitute back

u = y - 1

= ln (y - 1)^{2} + 1 + y - 1

Expand

(y - 1)^{2} + 1 : y^{2} - 2 y + 2

= ln y^{2} - 2 y + 2 + y - 1

Add a constant to the solution

= ln y^{2} - 2 y + 2 + y - 1 + C

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