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

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

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

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

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

Solution steps

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

b^{2} - x^{2} = b^{2} (1 - \frac{x ^{2}}{b ^{2}})

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

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{x ^{2}}{b ^{2}} = (\frac{x}{b ^{2}})^{2}

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

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

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

Take the constant out:

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

= \frac{1}{( b ^{2} ) ^{\frac{3}{2}}} \cdot \int \frac{1}{( 1 - ( \frac{x}{b ^{2}} ) ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int b sec^{2} (u) d u

= \frac{1}{( b ^{2} ) ^{\frac{3}{2}}} \cdot \int b sec^{2} (u) d u

Take the constant out:

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

= \frac{1}{( b ^{2} ) ^{\frac{3}{2}}} b \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= \frac{1}{( b ^{2} ) ^{\frac{3}{2}}} b tan (u)

Substitute back

u = arcsin (\frac{1}{b} x)

= \frac{1}{( b ^{2} ) ^{\frac{3}{2}}} b tan (arcsin (\frac{1}{b} x))

Simplify

\frac{1}{( b ^{2} ) ^{\frac{3}{2}}} b tan (arcsin (\frac{1}{b} x)) : \frac{x}{b ^{2} ( - x ^{2} + b ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

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\int 14 e^{x} d x

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