What is the integral of 1/((x^2+6x+10)^{3/2)} ?

The integral of 1/((x^2+6x+10)^{3/2)} is (x+3)/((x^2+6x+10)^{1/2)}+C

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integral of 1/((x^2+6x+10)^{3/2)}

\int \frac{1}{( x ^{2} + 6 x + 10 ) ^{\frac{3}{2}}} d x

\frac{x + 3}{( x ^{2} + 6 x + 10 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} + 6 x + 10 ) ^{\frac{3}{2}}} d x

Complete the square

x^{2} + 6 x + 10 : (x + 3)^{2} + 1

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int cos (v) d v

= \int cos (v) d v

Use the common integral:

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

= sin (v)

Substitute back

= sin (arctan (x + 3))

Simplify

sin (arctan (x + 3)) : \frac{x + 3}{( x ^{2} + 6 x + 10 ) ^{\frac{1}{2}}}

= \frac{x + 3}{( x ^{2} + 6 x + 10 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x + 3}{( x ^{2} + 6 x + 10 ) ^{\frac{1}{2}}} + C

\int \frac{1}{( 2 x + 1 ) 4 x ^{2} + 4 x} d x

\int \frac{sin ^{4} ( x )}{cos ( x )} d x

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

\int (e^{x} + sin (x))^{2} d x

\int \frac{x}{( x ^{2} + 2 x + 10 ) ^{3}} d x

What is the integral of 1/((x^2+6x+10)^{3/2)} ?
The integral of 1/((x^2+6x+10)^{3/2)} is (x+3)/((x^2+6x+10)^{1/2)}+C