Question

integration of x^2.tan^-1 x.dx​

Answer 1

EXPLANATION.

$\implies \displaystyle \int x^{2} tan^{-1} x dx$

As we know that,

If two functions is given in the integration then apply integration by-parts, we get.

If u and v are two functions of x, then.

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

From the first letter of the words,

I = Inverse trigonometric functions.

L = Logarithmic functions.

A = Algebraic functions.

T = Trigonometric functions.

E = Exponential functions.

We get a word = ILATE.

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

tan⁻¹x = first function.

x² = second function.

$\implies \displaystyle tan^{-1}x \int x^{2} dx \ - \int \bigg[ \dfrac{d(tan^{-1}x) }{dx} . \int x^{2} dx \bigg] dx$

$\implies \displaystyle tan^{-1} x . \dfrac{x^{3} }{3} \ - \int \bigg[ \dfrac{1}{1 + x^{2} } \dfrac{x^{3} }{3} \bigg] dx$

$\implies \displaystyle \dfrac{tan^{-1}x . x^{3} }{3} \ - \dfrac{1}{3} \int \dfrac{x^{3} }{1 + x^{2} } dx$

1 + x² = t.

Differentiate w.r.t x, we get.

⇒ 2xdx = dt.

⇒ xdx = dt/2.

⇒ x² = t - 1.

$\implies \displaystyle \dfrac{x^{3} }{3} tan^{-1} x \ - \dfrac{1}{6} \int \dfrac{t - 1}{t} dt$

$\implies \displaystyle \dfrac{x^{3} }{3} tan^{-1} x \ - \dfrac{1}{6} \bigg( \int \dfrac{-dt}{t} + \int dt \bigg)$

$\implies \displaystyle \dfrac{x^{3} }{3} tan^{-1} x \ - \dfrac{1}{6} (-ln (t) + t ) + C$

$\implies \displaystyle \dfrac{x^{3} }{3} tan^{-1} x \ - \dfrac{1}{6} \bigg( - ln(1 + x^{2} ) + (1 + x^{2} ) \bigg) + C$

Answer 2

$\rm :\longmapsto\:\displaystyle\int\tt {x}^{2} {tan}^{ - 1}x \: dx$

Using integration by parts,

$\rm \:= {tan}^{ - 1}x\displaystyle\int\tt {x}^{2}dx - \displaystyle\int\tt \bigg(\dfrac{d}{dx}{tan}^{ - 1}x\displaystyle\int\tt {x}^{2} dx \bigg)x$

$\rm \:  = {tan}^{ - 1}x \: \times \dfrac{ {x}^{3} }{3} - \displaystyle\int\tt \dfrac{1}{1 + {x}^{2} } \times \dfrac{ {x}^{3} }{3} \: dx$

$\rm \:  =  \:  \: \dfrac{ {x}^{3}{tan}^{ - 1}x }{3} - \dfrac{1}{3} \displaystyle\int\tt \dfrac{ {x}^{3} }{1 + {x}^{2} } dx$

$\rm \:  =  \:  \: \dfrac{ {x}^{3}{tan}^{ - 1}x }{3} - \dfrac{1}{3} \displaystyle\int\tt \dfrac{ {x}^{2} \times x }{1 + {x}^{2} } dx$

$\rm \:  =  \:  \: \dfrac{ {x}^{3}{tan}^{ - 1}x }{3} - \dfrac{1}{3} \displaystyle\int\tt \dfrac{( {x}^{2} + 1 - 1) \times x }{1 + {x}^{2} } dx$

$\rm \:  =  \:  \: \dfrac{ {x}^{3}{tan}^{ - 1}x }{3} - \dfrac{1}{3} \displaystyle\int\tt \dfrac{( {x}^{2} + 1)x - x }{1 + {x}^{2} } dx$

$\rm \:  =  \:  \: \dfrac{ {x}^{3}{tan}^{ - 1}x }{3} - \dfrac{1}{3} \displaystyle\int\tt \dfrac{( {x}^{2} + 1)x }{1 + {x}^{2} } dx + \dfrac{1}{3}\displaystyle\int\tt \dfrac{x}{ {x}^{2} + 1}dx$

$\rm \:  =  \:  \: \dfrac{ {x}^{3}{tan}^{ - 1}x }{3} - \dfrac{1}{3} \displaystyle\int\tt x dx + \dfrac{1}{6}\displaystyle\int\tt \dfrac{2x}{ {x}^{2} + 1}dx$

$\rm \:  =  \:  \: \dfrac{ {x}^{3}{tan}^{ - 1}x }{3} - \dfrac{1}{3} \times \dfrac{ {x}^{2} }{2} + \dfrac{1}{6}log | {x}^{2} + 1 | + c$

$\rm \:  =  \:  \: \dfrac{ {x}^{3}{tan}^{ - 1}x }{3} - \dfrac{ {x}^{2} }{6} + \dfrac{1}{6}log | {x}^{2} + 1 | + c$

Formula used to solve this question

Integration by Parts

Formula is

∫u v dx = u∫v dx −∫u' (∫v dx) dx

u is the function u(x)

v is the function v(x)

u' is the derivative of the function u(x)

For integration by parts , the ILATE rule is used to choose u and v.

where,

I - Inverse trigonometric functions

L -Logarithmic functions

A - Arithmetic and Algebraic functions

T - Trigonometric functions

E- Exponential functions

The alphabet which comes first is choosen as u and other as v.

$\boxed{ \sf{ \: \displaystyle\int\tt x \: dx = \dfrac{ {x}^{2} }{2} + c}}$

$\boxed{ \sf{ \: \displaystyle\int\tt \dfrac{f'(x)}{f(x)} dx = log |f(x)| + c}}$