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

The integral of 1/(1+(x-3)^2) is arctan(x-3)+C

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

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

arctan (x - 3) + C

Solution steps

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

Apply u-substitution

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

Use the common integral:

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

= arctan (u)

Substitute back

u = x - 3

= arctan (x - 3)

Add a constant to the solution

= arctan (x - 3) + C

What is the integral of 1/(1+(x-3)^2) ?
The integral of 1/(1+(x-3)^2) is arctan(x-3)+C