Question

(Vx2+1)(log(x2 +1)-2logx)
Evaluate: S
-dx
x4​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle\int \dfrac{\sqrt{x^{2} + 1} \bigg[log(x^{2} + 1) - 2logx\bigg]}{x^{4} } dx.$

As we know that,

We can write equation as,

$\sf \implies \displaystyle\int \dfrac{x \bigg(1 \ + \dfrac{1}{x^{2} } \bigg)^{\dfrac{1}{2} }\bigg[log(x^{2} + 1) - logx^{2} \bigg] }{x^{4} } dx.$

$\sf \implies \displaystyle\int \dfrac{\bigg(1 \ + \dfrac{1}{x^{2} } \bigg)^{\dfrac{1}{2} } \bigg[log \bigg(\dfrac{x^{2} + 1}{x^{2} } \bigg)\bigg]}{x^{3} } dx.$

$\sf \implies \displaystyle\int \dfrac{\bigg(1 \ + \dfrac{1}{x^{2} }\bigg)^{\dfrac{1}{2} }\bigg[log\bigg(1 \ + \dfrac{1}{x^{2} }\bigg)\bigg] }{x^{3} } dx.$

As we know that,

Apply substitution method in this questions, we get.

Let we assume that,

⇒ (1 + 1/x²) = t.

⇒ -2/x³.dx = dt.

⇒ -dt/2 = dx/x³.

Put the value in this equation, we get.

$\sf \implies \displaystyle\int (t)^{\dfrac{1}{2} } \ log (t)\ \dfrac{-dt}{2}$

$\sf \implies \dfrac{-1}{2} \displaystyle\int t^{\dfrac{1}{2} } \ log(t) dt.$

As we know that,

We can apply Integration by parts rules, we get.

$\sf \implies \displaystyle\int (u.v)dx = u \bigg(\int v dx\bigg) - \int \bigg[ \dfrac{du}{dx} . \bigg(\int v dx\bigg)\bigg]dx.$

I = Inverse trigonometric function.

L = Logarithmic function.

A = Algebraic function.

T = Trigonometric function.

E = Exponential function.

This is known as ILATE rule.

First arrange the functions in the order according to letters of this word and then integrate by parts.

⇒ ㏒(t) = 1st function.

⇒ t^1/2 = 2nd function.

$\sf \implies \dfrac{-1}{2} \bigg[log(t). \displaystyle\int t^{\dfrac{1}{2} }dt \ \bigg]- \int \bigg[\dfrac{d(log(t))}{dt} . \int t^{\dfrac{1}{2} }dt \bigg]dt$

As we know that,

Formula of :

⇒ ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ -1).

Apply this formula in equation, we get.

$\sf \implies \dfrac{-1}{2} \bigg[\dfrac{2}{3}t^{\dfrac{3}{2} } log (t) \ - \dfrac{2}{3} \displaystyle\int t^{\dfrac{1}{2} }dt \bigg]$

$\sf \implies \dfrac{-1}{2} \bigg[ \dfrac{2}{3} t^{\dfrac{3}{2} } log(t) \ - \dfrac{4}{9} t^{\dfrac{3}{2} } \bigg]$

$\sf \implies \dfrac{-1}{3} t^{\dfrac{3}{2} } log(t) \ + \dfrac{2}{9} t^{\dfrac{3}{2} }$

Put the value of t = 1 + 1/x² in equation, we get.

$\sf \implies \dfrac{-1}{3} \bigg(1 \ + \dfrac{1}{x^{2} } \bigg)^{\dfrac{3}{2} } \ log \bigg(1 + \dfrac{1}{x^{2} }\bigg) \ + \dfrac{2}{9} \bigg(1 \ + \dfrac{1}{x^{2} } \bigg)^{\dfrac{3}{2} } \ + C.$

Answer 2

EXPLANATION.

$\sf \implies \displaystyle\int \dfrac{\sqrt{x^{2} + 1} \bigg[log(x^{2} + 1) - 2logx\bigg]}{x^{4} } dx.$

As we know that,

We can write equation as,

$\sf \implies \displaystyle\int \dfrac{x \bigg(1 \ + \dfrac{1}{x^{2} } \bigg)^{\dfrac{1}{2} }\bigg[log(x^{2} + 1) - logx^{2} \bigg] }{x^{4} } dx.$

$\sf \implies \displaystyle\int \dfrac{\bigg(1 \ + \dfrac{1}{x^{2} } \bigg)^{\dfrac{1}{2} } \bigg[log \bigg(\dfrac{x^{2} + 1}{x^{2} } \bigg)\bigg]}{x^{3} } dx.$

$\sf \implies \displaystyle\int \dfrac{\bigg(1 \ + \dfrac{1}{x^{2} }\bigg)^{\dfrac{1}{2} }\bigg[log\bigg(1 \ + \dfrac{1}{x^{2} }\bigg)\bigg] }{x^{3} } dx.$

As we know that,

Apply substitution method in this questions, we get.

Let we assume that,

⇒ (1 + 1/x²) = t.

⇒ -2/x³.dx = dt.

⇒ -dt/2 = dx/x³.

Put the value in this equation, we get.

$\sf \implies \displaystyle\int (t)^{\dfrac{1}{2} } \ log (t)\ \dfrac{-dt}{2}$

$\sf \implies \dfrac{-1}{2} \displaystyle\int t^{\dfrac{1}{2} } \ log(t) dt.$

As we know that,

We can apply Integration by parts rules, we get.

$\sf \implies \displaystyle\int (u.v)dx = u \bigg(\int v dx\bigg) - \int \bigg[ \dfrac{du}{dx} . \bigg(\int v dx\bigg)\bigg]dx.$

I = Inverse trigonometric function.

L = Logarithmic function.

A = Algebraic function.

T = Trigonometric function.

E = Exponential function.

This is known as ILATE rule.

First arrange the functions in the order according to letters of this word and then integrate by parts.

⇒ ㏒(t) = 1st function.

⇒ t^1/2 = 2nd function.

$\sf \implies \dfrac{-1}{2} \bigg[log(t). \displaystyle\int t^{\dfrac{1}{2} }dt \ \bigg]- \int \bigg[\dfrac{d(log(t))}{dt} . \int t^{\dfrac{1}{2} }dt \bigg]dt$

As we know that,

Formula of :

⇒ ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ -1).

Apply this formula in equation, we get.

$\sf \implies \dfrac{-1}{2} \bigg[\dfrac{2}{3}t^{\dfrac{3}{2} } log (t) \ - \dfrac{2}{3} \displaystyle\int t^{\dfrac{1}{2} }dt \bigg]$

$\sf \implies \dfrac{-1}{2} \bigg[ \dfrac{2}{3} t^{\dfrac{3}{2} } log(t) \ - \dfrac{4}{9} t^{\dfrac{3}{2} } \bigg]$

$\sf \implies \dfrac{-1}{3} t^{\dfrac{3}{2} } log(t) \ + \dfrac{2}{9} t^{\dfrac{3}{2} }$

Put the value of t = 1 + 1/x² in equation, we get.

$\sf \implies \dfrac{-1}{3} \bigg(1 \ + \dfrac{1}{x^{2} } \bigg)^{\dfrac{3}{2} } \ log \bigg(1 + \dfrac{1}{x^{2} }\bigg) \ + \dfrac{2}{9} \bigg(1 \ + \dfrac{1}{x^{2} } \bigg)^{\dfrac{3}{2} } \ + C.$