What is the integral of 1/((z^2-6z+18)^{3/2)} ?

The integral of 1/((z^2-6z+18)^{3/2)} is (z-3)/(9(z^2-6z+18)^{1/2)}+C

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integral of 1/((z^2-6z+18)^{3/2)}

\int \frac{1}{( z ^{2} - 6 z + 18 ) ^{\frac{3}{2}}} d z

\frac{z - 3}{9 ( z ^{2} - 6 z + 18 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( z ^{2} - 6 z + 18 ) ^{\frac{3}{2}}} d z

Complete the square

z^{2} - 6 z + 18 : (z - 3)^{2} + 9

= \int \frac{1}{( ( z - 3 ) ^{2} + 9 ) ^{\frac{3}{2}}} d z

Apply u-substitution

: \int \frac{1}{( u ^{2} + 9 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( u ^{2} + 9 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int \frac{1}{9 sec ( v )} d v

= \int \frac{1}{9 sec ( v )} d v

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{sec ( v )} d v

Rewrite using trig identities

= \frac{1}{9} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= \frac{1}{9} sin (v)

Substitute back

= \frac{1}{9} sin (arctan (\frac{1}{3} (z - 3)))

Simplify

\frac{1}{9} sin (arctan (\frac{1}{3} (z - 3))) : \frac{z - 3}{9 ( z ^{2} - 6 z + 18 ) ^{\frac{1}{2}}}

= \frac{z - 3}{9 ( z ^{2} - 6 z + 18 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{z - 3}{9 ( z ^{2} - 6 z + 18 ) ^{\frac{1}{2}}} + C

x \to 0 lim (- 1)

e^{x^{7}}

x^{2} y^{^{''}} + x y^{^{'}} + 4 = 0

y^{^{''}} + 7 y^{^{'}} - 3 y = 0, y (0) = - 3, y^{^{'}} (0) = 7

\int \frac{2 x ^{\frac{3}{2}}}{3} d x

What is the integral of 1/((z^2-6z+18)^{3/2)} ?
The integral of 1/((z^2-6z+18)^{3/2)} is (z-3)/(9(z^2-6z+18)^{1/2)}+C