What is the integral of 1/((x^2+8x+25)^{3/2)} ?

The integral of 1/((x^2+8x+25)^{3/2)} is (x+4)/(9(x^2+8x+25)^{1/2)}+C

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integral of 1/((x^2+8x+25)^{3/2)}

\int \frac{1}{( x ^{2} + 8 x + 25 ) ^{\frac{3}{2}}} d x

\frac{x + 4}{9 ( x ^{2} + 8 x + 25 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} + 8 x + 25 ) ^{\frac{3}{2}}} d x

Complete the square

x^{2} + 8 x + 25 : (x + 4)^{2} + 9

= \int \frac{1}{( ( x + 4 ) ^{2} + 9 ) ^{\frac{3}{2}}} d x

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} (x + 4)))

Simplify

\frac{1}{9} sin (arctan (\frac{1}{3} (x + 4))) : \frac{x + 4}{9 ( x ^{2} + 8 x + 25 ) ^{\frac{1}{2}}}

= \frac{x + 4}{9 ( x ^{2} + 8 x + 25 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x + 4}{9 ( x ^{2} + 8 x + 25 ) ^{\frac{1}{2}}} + C

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

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\int e^{t + l n (t)} d t

What is the integral of 1/((x^2+8x+25)^{3/2)} ?
The integral of 1/((x^2+8x+25)^{3/2)} is (x+4)/(9(x^2+8x+25)^{1/2)}+C