What is the integral of 7/((1-x^2)^{3/2)} ?

The integral of 7/((1-x^2)^{3/2)} is (7x)/((1-x^2)^{1/2)}+C

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integral of 7/((1-x^2)^{3/2)}

\int \frac{7}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{7 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{7}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

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

= 7 \cdot \int sec^{2} (u) d u

Use the common integral:

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

= 7 tan (u)

Substitute back

u = arcsin (x)

= 7 tan (arcsin (x))

Simplify

7 tan (arcsin (x)) : \frac{7 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{7 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{7 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

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

x \to 1 lim (4 x^{3} - x^{2} + 7 x)

\frac{d}{d x} (\frac{1 + csc ( x )}{1 - csc ( x )})

\int cos^{2} (4 x) sin^{3} (4 x) d x

x \to \infty lim (7^{2 x})

What is the integral of 7/((1-x^2)^{3/2)} ?
The integral of 7/((1-x^2)^{3/2)} is (7x)/((1-x^2)^{1/2)}+C