How do you use substitution to integrate $\frac{4 x^{3}}{{(x^{2} + 1)}^{2}}$ $(4x^3)/( (x^2+1)^2)$ ?

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How do you use substitution to integrate $\frac{4 x^{3}}{{(x^{2} + 1)}^{2}}$ $(4x^3)/( (x^2+1)^2)$ ?

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Answer 1 · 2018-03-05T22:44:15

I tried this:

Explanation:

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Answer 2 · 2018-03-05T22:53:56

$\int \frac{4 x^{3}}{{(x^{2} + 1)}^{2}} d x = 2 ln (1 + x^{2}) + \frac{2}{1 + x^{2}} + C$ $int (4x^3)/(x^2+1)^2dx = 2ln (1+x^2) +2/(1+x^2)+C$

Explanation:

Substitute $x = tan t$ $x = tant$ , $d x = {sec}^{2} t$ $dx= sec^2t$

$\int \frac{4 x^{3}}{{(x^{2} + 1)}^{2}} d x = \int \frac{4 {tan}^{3} t {sec}^{2} t}{{({tan}^{2} t + 1)}^{2}} d t$ $int (4x^3)/(x^2+1)^2dx = int (4tan^3tsec^2t)/(tan^2t+1)^2dt$

Use now the trigonometric identity:

$1 + {tan}^{2} t = {sec}^{2} t$ $1+tan^2t = sec^2t$

$\int \frac{4 x^{3}}{{(x^{2} + 1)}^{2}} d x = \int \frac{4 {tan}^{3} t {sec}^{2} t}{{sec}^{4} t} d t$ $int (4x^3)/(x^2+1)^2dx = int (4tan^3tsec^2t)/sec^4tdt$

$\int \frac{4 x^{3}}{{(x^{2} + 1)}^{2}} d x = \int \frac{4 {tan}^{3} t}{{sec}^{2} t} d t$ $int (4x^3)/(x^2+1)^2dx = int (4tan^3t)/sec^2tdt$

$\int \frac{4 x^{3}}{{(x^{2} + 1)}^{2}} d x = \int \frac{4 {sin}^{3} t {cos}^{2}}{{cos}^{3} t} d t$ $int (4x^3)/(x^2+1)^2dx = int (4sin^3tcos^2)/cos^3tdt$

$\int \frac{4 x^{3}}{{(x^{2} + 1)}^{2}} d x = \int \frac{4 {sin}^{3} t}{cos t} d t$ $int (4x^3)/(x^2+1)^2dx = int (4sin^3t)/costdt$

$\int \frac{4 x^{3}}{{(x^{2} + 1)}^{2}} d x = 4 \int \frac{(1 - {cos}^{2} t) sin t}{cos t} d t$ $int (4x^3)/(x^2+1)^2dx = 4int ((1-cos^2t)sint)/costdt$

$\int \frac{4 x^{3}}{{(x^{2} + 1)}^{2}} d x = - 4 \int (\frac{1}{cos t} - cos t) d (cos t)$ $int (4x^3)/(x^2+1)^2dx = -4int (1/cost-cost) d(cost)$

$\int \frac{4 x^{3}}{{(x^{2} + 1)}^{2}} d x = - 4 ln | cos t | + 2 {cos}^{2} t + C$ $int (4x^3)/(x^2+1)^2dx = -4ln abs (cost) +2 cos^2t + C$

using the properties of logarithms:

$\int \frac{4 x^{3}}{{(x^{2} + 1)}^{2}} d x = 2 ln (\frac{1}{{cos}^{2} t}) + 2 {cos}^{2} t + C$ $int (4x^3)/(x^2+1)^2dx = 2ln (1/ cos^2t) +2 cos^2t + C$

To undo the substitution note that:

${cos}^{2} t = \frac{1}{{sec}^{2} t} = \frac{1}{1 + {tan}^{2} t} = \frac{1}{1 + x^{2}}$ $cos^2t = 1/sec^2t = 1/(1+tan^2t) = 1/(1+x^2)$

$\int \frac{4 x^{3}}{{(x^{2} + 1)}^{2}} d x = 2 ln (1 + x^{2}) + \frac{2}{1 + x^{2}} + C$ $int (4x^3)/(x^2+1)^2dx = 2ln (1+x^2) +2/(1+x^2)+C$

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How do you use substitution to integrate $\frac{4 x^{3}}{{(x^{2} + 1)}^{2}}$ $(4x^3)/( (x^2+1)^2)$ ?

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