What is the integral of 1/((x+4)sqrt(x^2+8x+12)) ?

The integral of 1/((x+4)sqrt(x^2+8x+12)) is 1/2 arcsec(1/2 (x+4))+C

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integral of 1/((x+4)sqrt(x^2+8x+12))

\int \frac{1}{( x + 4 ) x ^{2} + 8 x + 12} d x

\frac{1}{2} arcsec (\frac{1}{2} (x + 4)) + C

Solution steps

\int \frac{1}{( x + 4 ) x ^{2} + 8 x + 12} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 4} d u

= \int \frac{1}{u u ^{2} - 4} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= \int \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (x + 4))

Simplify

= \frac{1}{2} arcsec (\frac{1}{2} (x + 4))

Add a constant to the solution

= \frac{1}{2} arcsec (\frac{1}{2} (x + 4)) + C

\frac{d}{d x} (\frac{x ^{4}}{2})

\int_{- 1}^{4} (x + 3) d x

\frac{d}{d x} (- 7 e^{x})

\frac{d}{d s} (\frac{6}{( s - 2 ) ^{2} + 36})

\int \frac{1}{x ^{5} + x ^{4} + 2 x ^{3} + 2 x ^{2} + x + 1} d x

What is the integral of 1/((x+4)sqrt(x^2+8x+12)) ?
The integral of 1/((x+4)sqrt(x^2+8x+12)) is 1/2 arcsec(1/2 (x+4))+C