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

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

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

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

512 - \frac{1}{32 ( 64 x ^{2} + 48 x + 16 ) ^{2}} - \frac{3 ( 12288 arctan ( \frac{1}{7} ( 8 x + 3 ) ) ( x + \frac{3}{8} ) ^{4} + 1536 7 ( x + \frac{3}{8} ) ^{3} + 2688 arctan ( \frac{1}{7} ( 8 x + 3 ) ) ( x + \frac{3}{8} ) ^{2} + 280 7 ( x + \frac{3}{8} ) + 147 arctan ( \frac{1}{7} ( 8 x + 3 ) ) )}{3136 7 ( 64 x ^{2} + 48 x + 16 ) ^{2}} + C

Solution steps

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

Complete the square

4 x^{2} + 3 x + 1 : 4 (x + \frac{3}{8})^{2} + \frac{7}{16}

= \int \frac{x}{( 4 ( x + \frac{3}{8} ) ^{2} + \frac{7}{16} ) ^{3}} d x

Apply u-substitution

: \int \frac{512 ( 8 u - 3 )}{( 64 u ^{2} + 7 ) ^{3}} d u

= \int \frac{512 ( 8 u - 3 )}{( 64 u ^{2} + 7 ) ^{3}} d u

Take the constant out:

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

= 512 \cdot \int \frac{8 u - 3}{( 64 u ^{2} + 7 ) ^{3}} d u

Expand

\frac{8 u - 3}{( 64 u ^{2} + 7 ) ^{3}} : \frac{8 u}{( 64 u ^{2} + 7 ) ^{3}} - \frac{3}{( 64 u ^{2} + 7 ) ^{3}}

= 512 \cdot \int \frac{8 u}{( 64 u ^{2} + 7 ) ^{3}} - \frac{3}{( 64 u ^{2} + 7 ) ^{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 512 (\int \frac{8 u}{( 64 u ^{2} + 7 ) ^{3}} d u - \int \frac{3}{( 64 u ^{2} + 7 ) ^{3}} d u)

\int \frac{8 u}{( 64 u ^{2} + 7 ) ^{3}} d u = - \frac{1}{32 ( 64 u ^{2} + 7 ) ^{2}}

\int \frac{3}{( 64 u ^{2} + 7 ) ^{3}} d u = \frac{3 ( 12288 u ^{4} arctan ( \frac{8 u}{7} ) + 1536 7 u ^{3} + 2688 u ^{2} arctan ( \frac{8 u}{7} ) + 280 7 u + 147 arctan ( \frac{8 u}{7} ) )}{3136 7 ( 64 u ^{2} + 7 ) ^{2}}

= 512 - \frac{1}{32 ( 64 u ^{2} + 7 ) ^{2}} - \frac{3 ( 12288 u ^{4} arctan ( \frac{8 u}{7} ) + 1536 7 u ^{3} + 2688 u ^{2} arctan ( \frac{8 u}{7} ) + 280 7 u + 147 arctan ( \frac{8 u}{7} ) )}{3136 7 ( 64 u ^{2} + 7 ) ^{2}}

Substitute back

u = x + \frac{3}{8}

= 512 - \frac{1}{32 ( 64 ( x + \frac{3}{8} ) ^{2} + 7 ) ^{2}} - \frac{3 ( 12288 ( x + \frac{3}{8} ) ^{4} arctan ( \frac{8 ( x + \frac{3}{8} )}{7} ) + 1536 7 ( x + \frac{3}{8} ) ^{3} + 2688 ( x + \frac{3}{8} ) ^{2} arctan ( \frac{8 ( x + \frac{3}{8} )}{7} ) + 280 7 ( x + \frac{3}{8} ) + 147 arctan ( \frac{8 ( x + \frac{3}{8} )}{7} ) )}{3136 7 ( 64 ( x + \frac{3}{8} ) ^{2} + 7 ) ^{2}}

Simplify

512 - \frac{1}{32 ( 64 ( x + \frac{3}{8} ) ^{2} + 7 ) ^{2}} - \frac{3 ( 12288 ( x + \frac{3}{8} ) ^{4} arctan ( \frac{8 ( x + \frac{3}{8} )}{7} ) + 1536 7 ( x + \frac{3}{8} ) ^{3} + 2688 ( x + \frac{3}{8} ) ^{2} arctan ( \frac{8 ( x + \frac{3}{8} )}{7} ) + 280 7 ( x + \frac{3}{8} ) + 147 arctan ( \frac{8 ( x + \frac{3}{8} )}{7} ) )}{3136 7 ( 64 ( x + \frac{3}{8} ) ^{2} + 7 ) ^{2}} : 512 - \frac{1}{32 ( 64 x ^{2} + 48 x + 16 ) ^{2}} - \frac{3 ( 12288 arctan ( \frac{1}{7} ( 8 x + 3 ) ) ( x + \frac{3}{8} ) ^{4} + 1536 7 ( x + \frac{3}{8} ) ^{3} + 2688 arctan ( \frac{1}{7} ( 8 x + 3 ) ) ( x + \frac{3}{8} ) ^{2} + 280 7 ( x + \frac{3}{8} ) + 147 arctan ( \frac{1}{7} ( 8 x + 3 ) ) )}{3136 7 ( 64 x ^{2} + 48 x + 16 ) ^{2}}

= 512 - \frac{1}{32 ( 64 x ^{2} + 48 x + 16 ) ^{2}} - \frac{3 ( 12288 arctan ( \frac{1}{7} ( 8 x + 3 ) ) ( x + \frac{3}{8} ) ^{4} + 1536 7 ( x + \frac{3}{8} ) ^{3} + 2688 arctan ( \frac{1}{7} ( 8 x + 3 ) ) ( x + \frac{3}{8} ) ^{2} + 280 7 ( x + \frac{3}{8} ) + 147 arctan ( \frac{1}{7} ( 8 x + 3 ) ) )}{3136 7 ( 64 x ^{2} + 48 x + 16 ) ^{2}}

Add a constant to the solution

= 512 - \frac{1}{32 ( 64 x ^{2} + 48 x + 16 ) ^{2}} - \frac{3 ( 12288 arctan ( \frac{1}{7} ( 8 x + 3 ) ) ( x + \frac{3}{8} ) ^{4} + 1536 7 ( x + \frac{3}{8} ) ^{3} + 2688 arctan ( \frac{1}{7} ( 8 x + 3 ) ) ( x + \frac{3}{8} ) ^{2} + 280 7 ( x + \frac{3}{8} ) + 147 arctan ( \frac{1}{7} ( 8 x + 3 ) ) )}{3136 7 ( 64 x ^{2} + 48 x + 16 ) ^{2}} + C

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