Question

$\int\limits^1_02x+3/5x²+1 dx ,Evaluate the given integral expression:$

Answer 1

HELLO DEAR,

GIVEN:-
$\bold{\int\limits^1_0 (2x + 3)/(5x^2 + 1).dx}$

put (5x² + 1) = t
=> 10x.dx = dt
also, [x = 1 , t = 6] and [x = 0 , t = 1]

$\bold{\int\limits^6_1 [1/5(dt/t)] + \int\limits^1_0 3/5(x^2 + 1/5)}$

$\bold{1/5[log|t|]^6_1 + 3/(5*1/\sqrt{5})[tan^{-1}{\sqrt{5}x}]^1_0}$

$\bold{1/5[log6 - log1] + 3/\sqrt{5}[tan^{-1}{\sqrt{5}} - tan^{-1}{0}]}$

$\bold{1/5log6 + \{3/\sqrt{5}\}tan^{-1}\sqrt{5}}$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

we have to find the value of $\int\limits^1_0{\frac{2x+3}{5x^2+1}}\,dx$

Let f(x) = 5x² + 1
differentiate with respect to x,
f'(x) = 10x
so, we have to arrange numerator in such a way that it contains f'(x).
2x + 3 = 1/5{10x} + 3

now, $\int\limits^1_0{\frac{1/5(10x)+3}{5x^2+1}}\,dx$

=$\frac{1}{5}\int\limits^1_0{\frac{10x}{5x^2+1}}\,dx+\frac{3}{5}\int\limits^1_0{\frac{1}{x^2+(1/\sqrt{5})^2}}\,dx$

now use formula,
$\int{\frac{f'(x)}{f(x)}}\,dx=logf(x)$
$\int{\frac{1}{x^2+a^2}}=\frac{1}{a}tan^{-1}\frac{x}{a}$

so, $=\frac{1}{5}[log(5x^2+1)]^1_0+\frac{3}{5}\frac{1}{\frac{1}{\sqrt{5}}}[tan^{-1}\frac{x}{\frac{1}{\sqrt{5}}}]^1_0$

$=\frac{1}{5}log6+\frac{3}{\sqrt{5}}tan^{-1}\sqrt{5}$