What is the integral of (3x)/((1+9x^2)^4) ?

The integral of (3x)/((1+9x^2)^4) is -1/(18(1+9x^2)^3)+C

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integral of (3x)/((1+9x^2)^4)

\int \frac{3 x}{( 1 + 9 x ^{2} ) ^{4}} d x

- \frac{1}{18 ( 1 + 9 x ^{2} ) ^{3}} + C

Solution steps

\int \frac{3 x}{( 1 + 9 x ^{2} ) ^{4}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{( 1 + 9 x ^{2} ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{18 u ^{4}} d u

= 3 \cdot \int \frac{1}{18 u ^{4}} d u

Take the constant out:

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

= 3 \cdot \frac{1}{18} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= 3 \cdot \frac{1}{18} (- \frac{1}{3 u ^{3}})

Substitute back

u = 1 + 9 x^{2}

= 3 \cdot \frac{1}{18} (- \frac{1}{3 ( 1 + 9 x ^{2} ) ^{3}})

Simplify

3 \cdot \frac{1}{18} (- \frac{1}{3 ( 1 + 9 x ^{2} ) ^{3}}) : - \frac{1}{18 ( 1 + 9 x ^{2} ) ^{3}}

= - \frac{1}{18 ( 1 + 9 x ^{2} ) ^{3}}

Add a constant to the solution

= - \frac{1}{18 ( 1 + 9 x ^{2} ) ^{3}} + C

sin^{2} (t)

\int 8 x^{7} e^{8 x^{8}} d x

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What is the integral of (3x)/((1+9x^2)^4) ?
The integral of (3x)/((1+9x^2)^4) is -1/(18(1+9x^2)^3)+C