Notice that if u=x16,then u2=x13andu3=x12. Also we have that du=16x56dx=16u5dx. So that, ∫21+x13dx=∫6u5u31+u2du=6∫u81+u2 From here, we can simply use some polynomial division to simplify the quotient.Notice then that: u81+u2=u6−u4+u2−1+1u2+1 Thus, we have the folowing: 6∫u81+u2du=6(∫u6−u4+u2−1+1u2+1)du So we have the last result: 6(∫u6−u4+u2−1+1u2+1)du=6u77−6u55+6u33−6u+6arctan⁡u) 6x767−6x565+6x123−6x16+6arctan⁡x13

Question: "Solve. ∫x1+x3dx=?"

sagnuhh

Answered question

2020-10-25

Solve.

\int \frac{\sqrt{x}}{1 + \sqrt[3]{x}} d x = ?

Answer & Explanation

casincal

Skilled2020-10-26Added 82 answers

Notice that if

u = x^{\frac{1}{6}}, t h e n u^{2} = x^{\frac{1}{3}} and u^{3} = x^{\frac{1}{2}}

.
Also we have that

d u = \frac{1}{6 x^{\frac{5}{6}}} d x = \frac{1}{6 u^{5}} d x

.
So that,

\int \frac{\sqrt{2}}{1 + x^{\frac{1}{3}}} d x = \int 6 u^{5} \frac{u^{3}}{1 + u^{2}} d u = 6 \int \frac{u^{8}}{1 + u^{2}}

From here, we can simply use some polynomial division to simplify the quotient.Notice then that:

\frac{u^{8}}{1 + u^{2}} = u^{6} - u^{4} + u^{2} - 1 + \frac{1}{u^{2} + 1}

Thus, we have the folowing:

6 \int \frac{u^{8}}{1 + u^{2}} d u = 6 (\int u^{6} - u^{4} + u^{2} - 1 + \frac{1}{u^{2} + 1}) d u

So we have the last result:

6 (\int u^{6} - u^{4} + u^{2} - 1 + \frac{1}{u^{2} + 1}) d u = \frac{6 u^{7}}{7} - \frac{6 u^{5}}{5} + \frac{6 u^{3}}{3} - 6 u + 6 \arctan u)

\frac{6 x^{\frac{7}{6}}}{7} - \frac{6 x^{\frac{5}{6}}}{5} + \frac{6 x^{\frac{1}{2}}}{3} - 6 x^{\frac{1}{6}} + 6 \arctan x^{\frac{1}{3}}

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