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

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

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

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

\frac{1}{3} arctan (\frac{x - 2}{3}) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 + u ^{2}} d u

= \int \frac{1}{3 + u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{3 ( 1 + v ^{2} )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

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

= \frac{1}{3} arctan (v)

Substitute back

= \frac{1}{3} arctan (\frac{x - 2}{3})

Add a constant to the solution

= \frac{1}{3} arctan (\frac{x - 2}{3}) + C

\frac{x ^{4}}{8} + \frac{1}{4 x ^{2}}

\int_{0}^{1} ln (1 + x^{2}) d x

1 + x, 1

\int \frac{1}{4} cos^{2} (2 x) d x

y^{^{'}} + 0.2 \cdot y = 5

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