What is the integral of 2/((25+x^2)^{3/2)} ?

The integral of 2/((25+x^2)^{3/2)} is (2x)/(25(25+x^2)^{1/2)}+C

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integral of 2/((25+x^2)^{3/2)}

\int \frac{2}{( 25 + x ^{2} ) ^{\frac{3}{2}}} d x

\frac{2 x}{25 ( 25 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{2}{( 25 + 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

= 2 \cdot \int \frac{1}{( 25 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{25 sec ( u )} d u

= 2 \cdot \int \frac{1}{25 sec ( u )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{25} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= 2 \cdot \frac{1}{25} \cdot \int cos (u) d u

Use the common integral:

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

= 2 \cdot \frac{1}{25} sin (u)

Substitute back

u = arctan (\frac{1}{5} x)

= 2 \cdot \frac{1}{25} sin (arctan (\frac{1}{5} x))

Simplify

2 \cdot \frac{1}{25} sin (arctan (\frac{1}{5} x)) : \frac{2 x}{25 ( 25 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{2 x}{25 ( 25 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{2 x}{25 ( 25 + x ^{2} ) ^{\frac{1}{2}}} + C

\int \frac{1}{6} sin (6 x) d x

y^{^{''}} + a y^{^{'}} + b y = 0

\int x a^{2} - x^{2} d x

f (x) = (x^{2} \cdot 7^{x})^{7}

x \to \infty lim (\frac{sin ( 2 x )}{x})

What is the integral of 2/((25+x^2)^{3/2)} ?
The integral of 2/((25+x^2)^{3/2)} is (2x)/(25(25+x^2)^{1/2)}+C