What is the integral of 1/(25+(x-6)^2) ?

The integral of 1/(25+(x-6)^2) is 1/5 arctan((x-6)/5)+C

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integral of 1/(25+(x-6)^2)

\int \frac{1}{25 + ( x - 6 ) ^{2}} d x

\frac{1}{5} arctan (\frac{x - 6}{5}) + C

Solution steps

\int \frac{1}{25 + ( x - 6 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{25 + u ^{2}} d u

= \int \frac{1}{25 + u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{5 ( 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}{5} \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}{5} arctan (v)

Substitute back

= \frac{1}{5} arctan (\frac{x - 6}{5})

Add a constant to the solution

= \frac{1}{5} arctan (\frac{x - 6}{5}) + C

\frac{1}{5 s ^{2} + 0.2 s}

\frac{\partial}{\partial y} (e^{5 y} cos (- 5 x))

\int e^{x} sin (4 x) d x

\frac{d}{d x} (lo g_{3} (4 x + 1))

\frac{\partial}{\partial y} (- 9 sin (3 x + yz))

What is the integral of 1/(25+(x-6)^2) ?
The integral of 1/(25+(x-6)^2) is 1/5 arctan((x-6)/5)+C