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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{3 sec ( u )} d u

= \int \frac{1}{3 sec ( u )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{3} sin (u)

Substitute back

u = arctan (\frac{1}{3 ^{\frac{1}{2}}} x)

= \frac{1}{3} sin (arctan (\frac{1}{3 ^{\frac{1}{2}}} x))

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{9 x ^{2} + 36 x + 35} d x

y = \frac{8 x}{x ^{2} - 1}, (3, 3)

y^{^{''}} + 625 y = sec (25 x)

e^{5 x}

\int e^{- 1 + x} d x

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