What is the integral of 1/(64+(x-1)^2) ?

The integral of 1/(64+(x-1)^2) is 1/8 arctan((x-1)/8)+C

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integral of 1/(64+(x-1)^2)

\int \frac{1}{64 + ( x - 1 ) ^{2}} d x

\frac{1}{8} arctan (\frac{x - 1}{8}) + C

Solution steps

\int \frac{1}{64 + ( x - 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{64 + u ^{2}} d u

= \int \frac{1}{64 + u ^{2}} d u

Apply Integral Substitution

= \int \frac{1}{8 ( 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}{8} \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}{8} arctan (v)

Substitute back

= \frac{1}{8} arctan (\frac{x - 1}{8})

Add a constant to the solution

= \frac{1}{8} arctan (\frac{x - 1}{8}) + C

\frac{d y}{d x} = 50 y - \frac{1}{10} y^{2}

x^{2} + 1

\int (\frac{1}{5 x ^{3.3}}) d x

\int \frac{12}{( 36 x ^{2} + 1 ) ^{2}} d x

ln (3 x^{2})

What is the integral of 1/(64+(x-1)^2) ?
The integral of 1/(64+(x-1)^2) is 1/8 arctan((x-1)/8)+C