Question

Unit IV
8. (a) Evaluate the following integral :
√xlog (1+x) dx​

Answer 1

Basic Concept Used :-

Integration by Parts

Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways.

See the rule:

• ∫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.

$\large\underline{\sf{Solution-}}$

$\sf \longmapsto \displaystyle\int \tt\: \sqrt{x} \: log(1 + x) dx$

As we know that,

By using substitution method, we get.

$\rm :\longmapsto\: \sqrt{x} = y$

$\rm :\implies\:x = {y}^{2}$

$\rm :\longmapsto\:Differentiate \: w.r.t \: x, we \: get$

$\rm :\longmapsto\:dx = 2y \: dy$

Substituting all these values in given integral, we get

$= \sf \displaystyle\int \tt \: y log(1 + {y}^{2} ) 2ydy$

$= 2\sf \displaystyle\int \tt \: {y}^{2} log(1 + {y}^{2} ) \: dy$

Now, Integrating using by parts,

Here,

• v = y²

• u = log(1 + y²)

$= \tt \: 2 log(1 + {y}^{2} ) \sf \displaystyle\int \tt \: {y}^{2} dy - 2 \sf \displaystyle\int \tt \:\bigg( \dfrac{d}{dy} log(1 + {y}^{2} ) \sf \displaystyle\int \tt \: {y}^{2}dy \bigg) dy$

$\rm \:= \dfrac{2}{3} log(1 + {y}^{2} ) {y}^{3} - 2 \sf \displaystyle\int \tt \:\dfrac{2y}{1 + {y}^{2} } \times \dfrac{ {y}^{3} }{3}$

$\: \: \: \: \: \: \: \because \: \boxed{ \bf \: \sf \displaystyle\int \tt \: {x}^{2}dx = \dfrac{ {x}^{3} }{3} \: \: \: and \: \: \: \dfrac{d}{dx}logx = \dfrac{1}{x} }$

$\rm \: = \dfrac{2 {y}^{3} }{3} log(1 + {y}^{2} ) - \dfrac{4}{3} \sf \displaystyle\int \tt \:\dfrac{ {y}^{4} }{1 + {y}^{2} } dy$

$\rm \: = \dfrac{2 {y}^{3} }{3} log(1 + {y}^{2} ) - \dfrac{4}{3} \sf \displaystyle\int \tt \:\dfrac{ {y}^{4} - 1 + 1 }{1 + {y}^{2} } dy$

$\rm \: = \dfrac{2 {y}^{3} }{3} log(1 + {y}^{2} ) - \dfrac{4}{3} \sf \displaystyle\int \tt \:\dfrac{ {y}^{4} - 1 }{1 + {y}^{2}} dy \: - \dfrac{4}{3} \sf \displaystyle\int \tt \:\dfrac{dy}{ {y}^{2} + 1 }$

$\rm \: = \: \dfrac{ {2y}^{3} }{3} log(1 + {y}^{2} ) - \dfrac{4}{3} \sf \displaystyle\int \tt \:( {y}^{2} - 1)dy - \dfrac{4}{3} {tan }^{ - 1}y + c$

$\rm \: = \: \dfrac{ {2y}^{3} }{3} log(1 + {y}^{2} ) - \dfrac{4}{3} \bigg(\dfrac{ {y}^{3} }{3} - y \bigg) - \dfrac{4}{3} {tan }^{ - 1}y + c$

$\rm \: = \: \dfrac{ {2y}^{3} }{3} log(1 + {y}^{2} ) - \dfrac{4 {y}^{3} }{9} + \dfrac{4y}{3} - \dfrac{4}{3} {tan }^{ - 1}y + c$

( On substituting back the value of y, we get)

$\rm \: = \: \dfrac{ {2( \sqrt{x}) }^{3} }{3} log(1 + x) - \dfrac{4 {( \sqrt{x} )}^{3} }{9} + \dfrac{4 \sqrt{x} }{3} - \dfrac{4}{3} {tan }^{ - 1} \sqrt{x} + c$

$\rm \: = \: \dfrac{ {{2x \sqrt{x} }}}{3} log(1 + x ) - \dfrac{4 x \sqrt{x} }{9} + \dfrac{4 \sqrt{x} }{3} - \dfrac{4}{3} {tan }^{ - 1}\sqrt{x} + c$