If I (x) = ∫ x² ( x)² dx and 1 (1) = 0 , then I (x) =

Correct Answer is : x327[9(log⁡x)2−6log⁡x+2]−227{x³}{27}≤ft[9( x)²-6 x+2]-2/2727x3[9(logx)2−6logx+2]−272

{x³}{18}≤ft[8( x)²-3 x]+7/18

{x³}{27}≤ft[9( x)²+6 x]-2/27

{x³}{27}≤ft[9( x)²-6 x+2]-2/27

{x³}{27}≤ft[9( x)²-6 x-2]+2/27

Correct Answer is : x327[9(log⁡x)2−6log⁡x+2]−227{x³}{27}≤ft[9( x)²-6 x+2]-2/2727x3[9(logx)2−6logx+2]−272

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AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

If I (x) = ∫

x^2 (\log x)^2

dx and 1 (1) = 0 , then I (x) =

Solution:

Consider the integral,

I(x)=\int x^{2}(\log x)^{2}\,dx

Simplify the above using "by parts".

I(x)\;=\frac{x^{3}}{3}(\log x)^{2}-\int\frac{x^{3}}{3}\;\frac{2\log x}{x}\,dx

\;=\frac{x^{3}}{3}(\log x)^{2}-\frac{2}{3}\left[\frac{x^{3}}{3}\right](\log x)-\int\frac{x^{3}}{3}\;\frac{1}{x}\,dx

\;=\frac{x^{3}}{3}(\log x)^{2}-\frac{2}{3}\left[\frac{x^{3}}{3}(\log x)-\frac{1}{3}\cdot\frac{x^{3}}{3}\right]+C

\;=\frac{x^{3}}{27}\left[9(\log x)^{2}-6(\log x)+2\right]+C

For I (1) =0

\frac{2}{27}+C=0

C=-\frac{2}{27}

This implies

I(x)=\frac{x^{3}}{27}\left[9(\log x)^{2}-6(\log x)+2\right]-\frac{2}{27}

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