What is the integral of 1/((7+6x-x^2)^{3/2)} ?

The integral of 1/((7+6x-x^2)^{3/2)} is (x-3)/(16(-x^2+6x+7)^{1/2)}+C

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integral of 1/((7+6x-x^2)^{3/2)}

\int \frac{1}{( 7 + 6 x - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x - 3}{16 ( - x ^{2} + 6 x + 7 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 7 + 6 x - x ^{2} ) ^{\frac{3}{2}}} d x

Complete the square

7 + 6 x - x^{2} : - (x - 3)^{2} + 16

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

Apply u-substitution

: \int \frac{1}{( - u ^{2} + 16 ) ^{\frac{3}{2}}} d u

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

Apply Trigonometric Substitution

: \int \frac{1}{16 cos ^{2} ( v )} d v

= \int \frac{1}{16 cos ^{2} ( v )} d v

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{1}{cos ^{2} ( v )} d v

Use the common integral:

\int \frac{1}{cos ^{2} ( v )} d v = tan (v)

= \frac{1}{16} tan (v)

Substitute back

= \frac{1}{16} tan (arcsin (\frac{1}{4} (x - 3)))

Simplify

\frac{1}{16} tan (arcsin (\frac{1}{4} (x - 3))) : \frac{x - 3}{16 ( - x ^{2} + 6 x + 7 ) ^{\frac{1}{2}}}

= \frac{x - 3}{16 ( - x ^{2} + 6 x + 7 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x - 3}{16 ( - x ^{2} + 6 x + 7 ) ^{\frac{1}{2}}} + C

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What is the integral of 1/((7+6x-x^2)^{3/2)} ?
The integral of 1/((7+6x-x^2)^{3/2)} is (x-3)/(16(-x^2+6x+7)^{1/2)}+C