Question

Evaluate the integrals: $\int^1_0 {\frac{x}{x^2+1}} \, dx$

Answer 1

Answer

$log\sqrt{2}$

Step-by-step explanation:

concept:

$\int{\frac{1}{x}}\:dx=logx+c$

I have applied substitution method (change of variable) to solve this integraal.

In substitution method, the given integrand(non integrable function) is changed into an integrable function by taking suitable substitution.

Let

$I=\int\limits^{1}_{0}{\frac{1}{x^2+1}}\:xdx$

Take

$t=x^2+1\\\\\frac{dt}{dx}=2x\\\\\frac{dt}{2}=x.dx$

when x=0, t=0+1=1

when x=1, t=1+1=2

Now,

$I=\int\limits^{2}_{1}{\frac{1}{t}}\:\frac{dt}{2}\\\\I=\frac{1}{2}[logt]^{2}_{1}\\\\I=\frac{1}{2}[log2-log1]\\\\I=log2^{\frac{1}{2}}\\\\I=log\sqrt{2}$