How do you integrate $\int \frac{x^3}{1+x^2}dx$

Question

$$\int \frac{x^3}{1+x^2}dx$$

what I did was,

$t = x^4$

$\displaystyle \frac{dt}{dx} = 4x^3$

$4 \displaystyle \int \frac{1}{1+\sqrt t}dt$

$4[2(\sqrt t - \ln(\sqrt t +1)]$

$8x^2 - 8\ln(x^2 +1)$

The answer given in my book is,

$\displaystyle \frac{x^2}{2} - \frac{\ln(x^2 + 1)}{2}$

Any help would be appreciated.

Answer 1

apanpapan3's comment explains your error, but Piano Land's recommendation also works. Since $u=x^2\implies du = 2x dx$, your integral is $$\frac{u du}{2(1+u)}=\frac{1}{2}\int(1-\frac{1}{1+u})du=\frac{u-\ln (1+u)}{2}+C=\frac{x^2-\ln (1+x^2)}{2}+C.$$

Answer 2

Note that $$\frac{x^3}{1+x^2}=x-\frac{x}{1+x^2}$$ the rest is easy. And you can write

$$\frac{x}{1+x^2}=\frac{1}{2}\frac{2x}{1+x^2}$$

Answer 3

$$\int \frac { x^{ 3 } }{ 1+x^{ 2 } } dx=\frac { 1 }{ 2 } \int \frac { { x }^{ 2 } }{ 1+x^{ 2 } } { d\left( { x }^{ 2 } \right) }=\\ =\frac { 1 }{ 2 } \int \frac { { x }^{ 2 }+1-1 }{ 1+x^{ 2 } } { d\left( { x }^{ 2 } \right) }=\frac { 1 }{ 2 } \left( \int { d\left( { x }^{ 2 } \right) -\frac { 1 }{ 2 } \int \frac { d\left( 1+{ x }^{ 2 } \right) }{ 1+x^{ 2 } } } \right) =\\ =\frac { 1 }{ 2 } \left( { x }^{ 2 }-\ln { \left( 1+{ x }^{ 2 } \right) } \right) +C$$