Question

Solve this integration

Answer 1

Topic:

Integration

Solution:

We need to evaluate the following integral.

$\displaystyle\int \dfrac{dx}{x^{\frac12} + x^{\frac14}}$

Since the LCM of 1/2 and 1/4 is 1/4, we use the method of substitution.

Put x = t⁴ so that dx = 4 t³ dt.

The above integral changes to,

$\displaystyle\longrightarrow\int \dfrac{4t^3}{(t^4)^{\frac12} + (t^4)^{\frac14}}\ dt$

$\displaystyle\longrightarrow\int \dfrac{4t^3}{t^2 + t}\ dt$

$\displaystyle\longrightarrow\int \dfrac{4t^3}{t(t + 1)}\ dt$

$\displaystyle\longrightarrow\int \dfrac{4t^2}{t + 1}\ dt$

$\displaystyle\longrightarrow\int \dfrac{4t^2 -4+ 4}{t + 1}\ dt$

$\displaystyle\longrightarrow\int \dfrac{4(t^2-1)+ 4}{t + 1}\ dt$

$\displaystyle\longrightarrow\int \dfrac{4(t+1)(t-1)+ 4}{t + 1}\ dt$

$\displaystyle\longrightarrow\int \dfrac{4(t+1)(t-1)}{t + 1} + \dfrac{4}{t+1}\ dt$

$\displaystyle\longrightarrow\int 4(t-1)+ \dfrac{4}{t+1}\ dt$

$\displaystyle\longrightarrow\int 4t-4\ dt+ 4\int\dfrac{1}{t+1}\ dt$

Now here, we can use the formulas,

$\boxed{\int \dfrac{1}{x+a}\, dx = \log|x+a| +C}$

$\boxed{\int x^n\ dx = \dfrac{x^{n+1}}{n+1} +C}$

Using the above rules here, we get:

${\displaystyle\longrightarrow4\ \dfrac{t^{1+1}}{1+1}-4\ \dfrac{t^{0+1}}{ 0 + 1}\ + 4\log|t+1| + C}$

$\displaystyle\longrightarrow4\ \dfrac{t^2}{2}-4t + 4\log|t+1| + C$

$\displaystyle\longrightarrow2 t^2-4t + 4\log|t+1| + C$

Now, undoing our substitution, we get:

${\displaystyle\longrightarrow2 (x^{1/4})^2-4(x^{1/4}) + 4\log|x^{1/4}+1| + C}$

$\displaystyle\longrightarrow2x^{1/2}-4x^{1/4} + 4\log|x^{1/4}+1| + C$

$\displaystyle\longrightarrow2\sqrt{x}-4\sqrt[4]{x}+ 4\log|\sqrt[4]{x}+1| + C$

Hence this is the required answer.