What is the integral of 1/(9+(x-1)^2) ?

The integral of 1/(9+(x-1)^2) is 1/3 arctan((x-1)/3)+C

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integral of 1/(9+(x-1)^2)

\int \frac{1}{9 + ( x - 1 ) ^{2}} d x

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

Solution steps

\int \frac{1}{9 + ( x - 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{9 + u ^{2}} d u

= \int \frac{1}{9 + u ^{2}} d u

Apply Integral Substitution

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

Use the common integral:

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

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

Substitute back

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

Add a constant to the solution

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

f (x) = 3 x^{2}, (- 2, 12)

\int 8 x^{6} y^{3} + 15 x^{5} y^{4} d y

\frac{d}{d x} (x^{4} e^{2 x})

\frac{1}{2} x^{2} + 8 x + 9

y^{^{'}} = y (x y^{2} + 4)

What is the integral of 1/(9+(x-1)^2) ?
The integral of 1/(9+(x-1)^2) is 1/3 arctan((x-1)/3)+C