What is the integral of 5/((x^2+1)^{3/2)} ?

The integral of 5/((x^2+1)^{3/2)} is (5x)/((1+x^2)^{1/2)}+C

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integral of 5/((x^2+1)^{3/2)}

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

\frac{5 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

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

Take the constant out:

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

= 5 \cdot \int \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int cos (u) d u

= 5 \cdot \int cos (u) d u

Use the common integral:

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

= 5 sin (u)

Substitute back

u = arctan (x)

= 5 sin (arctan (x))

Simplify

5 sin (arctan (x)) : \frac{5 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{5 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{5 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

\int \frac{e ^{8 x}}{e ^{8 x} + 2} d x

\frac{2}{s ^{2} + 4}

\frac{d}{d x} (I n (x + a^{2} + x^{2}))

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

y^{^{'}} + 4 y = e^{- 4 t}

What is the integral of 5/((x^2+1)^{3/2)} ?
The integral of 5/((x^2+1)^{3/2)} is (5x)/((1+x^2)^{1/2)}+C