What is the integral of 7/(x^2-6x+25) ?

The integral of 7/(x^2-6x+25) is 7/4 arctan(1/4 (x-3))+C

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integral of 7/(x^2-6x+25)

\int \frac{7}{x ^{2} - 6 x + 25} d x

\frac{7}{4} arctan (\frac{1}{4} (x - 3)) + C

Solution steps

\int \frac{7}{x ^{2} - 6 x + 25} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{x ^{2} - 6 x + 25} d x

Complete the square

x^{2} - 6 x + 25 : (x - 3)^{2} + 16

= 7 \cdot \int \frac{1}{( x - 3 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 16} d u

= 7 \cdot \int \frac{1}{u ^{2} + 16} d u

Apply Integral Substitution

= 7 \cdot \int \frac{1}{4 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 7 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 7 \cdot \frac{1}{4} arctan (v)

Substitute back

= 7 \cdot \frac{1}{4} arctan (\frac{x - 3}{4})

Simplify

7 \cdot \frac{1}{4} arctan (\frac{x - 3}{4}) : \frac{7}{4} arctan (\frac{1}{4} (x - 3))

= \frac{7}{4} arctan (\frac{1}{4} (x - 3))

Add a constant to the solution

= \frac{7}{4} arctan (\frac{1}{4} (x - 3)) + C

x^{2} + 2 x, 1, 2

\int_{- 2}^{2} \frac{16}{x ^{2} - 4 x - 60} d x

y^{^{'''}} + y^{^{'}} = x

\frac{d}{d x} ((x - 1) e^{x})

\frac{7}{s ^{2} + 42}

What is the integral of 7/(x^2-6x+25) ?
The integral of 7/(x^2-6x+25) is 7/4 arctan(1/4 (x-3))+C