If (1+x) (1+x²) \, dx = (F(x) (1+x²) - 2/3 ^{-1} x - {2x³}{9} - {x²}{2} + 2x/3 + c) then, F(x) =

Correct Answer is : x22+x33+12{x²}{2} + {x³}{3} + 1/22x2+3x3+21

Correct Answer is : x22+x33+12{x²}{2} + {x³}{3} + 1/22x2+3x3+21

Test Index

TS EAMCET 5 May 2018 Shift 2 Solved Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

\int (1+x) \log (1+x^{2}) \, dx =

(F(x) \log (1+x^{2}) - \frac{2}{3} \tan^{-1} x -

\frac{2x^{3}}{9} - \frac{x^{2}}{2} + \frac{2x}{3} + c)

F(x) =

Solution:

Consider the integral.

\int (1+x) \log (1+x^{2}) \, dx =

(F(x) \log (1+x^{2}) - \frac{2}{3} \tan^{-1} x -

\frac{2x^{3}}{9} - \frac{x^{2}}{2} + \frac{2x}{3} + c)

Let,

I = \int (1+x) \log (1+x^{2}) \, dx

= \int (x + x^{2}) \log (1+x^{2}) \, dx

Solve the above integration by parts,

I = \log (1+x^{2}) \left( \frac{x^{2}}{2} + \frac{x^{3}}{3} \right)

- \int \frac{1}{1+x^{2}} \cdot 2x \left( \frac{x^{2}}{2} + \frac{x^{3}}{3} \right) \, dx

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

- \frac{2}{3} \int \left( \frac{x^{4}}{1+x^{2}} \right) \, dx

Let

I_{1} = \int \frac{x \cdot x^{2}}{1+x^{2}} \, dx

Let

x^{2} = t

2x \, dx = dt

I_{1} = \frac{1}{2} \int \frac{t}{1+t} \, dt

= \frac{1}{2} t - \frac{1}{2} \log |1+t| + c_{1}

= \frac{1}{2} x^{2} - \frac{1}{2} \log (1+x^{2}) + c_{1}

Let

I_{2} = \int \frac{x^{4}}{1+x^{2}} \, dx

\int \frac{x^{4}}{1+x^{2}} \, dx = \int \frac{x^{4} - 1 + 1}{1+x^{2}} \, dx

= \int \left( \frac{x^{2}-1}{1+x^{2}} + \frac{1}{1+x^{2}} \right) \, dx

= \frac{x^{3}}{3} - x + \tan^{-1} x + c_{2}

Therefore,

I = \left[ \log (1+x^{2}) \left( \frac{x^{2}}{2} + \frac{x^{3}}{3} \right) \right]

- \frac{1}{2} x^{2} + \frac{1}{2} \log (1+x^{2}) -

\left[ \frac{2}{9} x^{3} + \frac{2}{3} x - \frac{2}{3} \tan^{-1} x + c \right]

I = \left[ \log (1+x^{2}) \left( \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{1}{2} \right) \right]

- \frac{2}{3} \tan^{-1} x - \frac{2}{9} x^{3} -

\left[\frac{x^{2}}{2} + \frac{2}{3} x + C\right]

Comparing with original equation, we get

F(x) = \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{1}{2}

Go to Question:

Prev Question

Next Question