Advertisement Remove all ads

∫ 5 ( X 2 + 1 ) ( X + 2 ) D X - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum
\[\int\frac{5}{\left( x^2 + 1 \right) \left( x + 2 \right)} dx\]
Advertisement Remove all ads

Solution

We have,

\[I = \int\frac{5 dx}{\left( x^2 + 1 \right) \left( x + 2 \right)}\]

\[\text{Let }\frac{5}{\left( x + 2 \right) \left( x^2 + 1 \right)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 1}\]

\[ \Rightarrow \frac{5}{\left( x + 2 \right) \left( x^2 + 1 \right)} = \frac{A \left( x^2 + 1 \right) + \left( Bx + C \right) \left( x + 2 \right)}{\left( x + 2 \right) \left( x^2 + 1 \right)}\]

\[ \Rightarrow 5 = A \left( x^2 + 1 \right) + B x^2 + 2Bx + Cx + 2C\]

\[ \Rightarrow 5 = \left( A + B \right) x^2 + \left( 2B + C \right) x + \left( A + 2C \right)\]

\[\text{Equating coefficients of like terms}\]

\[A + B = 0 . . . . . \left( 1 \right)\]

\[2B + C = 0 . . . . . \left( 2 \right)\]

\[A + 2C = 5 . . . . . \left( 3 \right)\]

\[\text{Solving (1), (2) and (3), we get}\]

\[A = 1\]

\[B = - 1\]

\[C = 2\]

\[ \therefore \frac{5}{\left( x + 2 \right) \left( x^2 + 1 \right)} = \frac{1}{x + 2} + \left( \frac{- x + 2}{x^2 + 1} \right)\]

\[ \Rightarrow \int\frac{5 dx}{\left( x + 2 \right) \left( x^2 + 1 \right)} = \int\frac{dx}{x + 2} - \int\frac{x dx}{x^2 + 1} + 2\int\frac{dx}{x^2 + 1}\]

\[\text{Let }x^2 + 1 = t\]

\[ \Rightarrow 2xdx = dt\]

\[ \Rightarrow x dx = \frac{dt}{2}\]

\[ \therefore I = \int\frac{dx}{x + 2} - \frac{1}{2}\int\frac{dt}{t} + 2\int\frac{dx}{x^2 + 1^2}\]

\[ = \log \left| x + 2 \right| - \frac{1}{2} \log \left| t \right| + 2 \tan^{- 1} x + C'\]

\[ = \log \left| x + 2 \right| - \frac{1}{2} \log \left| x^2 + 1 \right| + 2 \tan^{- 1} x + C'\]

Concept: Indefinite Integral Problems
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 36 | Page 177

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×