Answer :
Answer:
1.5 * ln2
Step-by-step explanation:
Just multiply and divide the integrand by (2/3).
So your integral now becomes :
[tex]\frac{3}{2} \int\limits^1_0 {\frac{2x}{1+x^{2}} } \, dx[/tex]
Let t = 1 + x^2
Hence dt = 2x dx
Changing the limits :
When x = 0, t = 1 + 0 = 1
When x = 1, t = 1 + 1 = 2
2x dx is the numerator, which is dt and 1 + x^2 is the denominator, which is t.
So we have :
[tex]\frac{3}{2} \int\limits^2_1 {\frac{dt}{t}} \,[/tex]
Which is 1.5 * (ln(|t|)] from 1 to 2
= 1.5 * (ln|2| - ln|1|)
= 1.5 * ln2
Hence, the answer to your question is 3ln(2)/2, Which is approximately 1.04.
Hope you have understood my explanation.
Have a great day ahead !