logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Find the integral $\int\frac{\large x^3+5x^2-4}{\large x^2}\;dx$

$\begin{array}{1 1} \frac{x^2}{2}+5x+\frac{4}{x}+c \\ \frac{x^3}{2}+5x^2+ \frac{4}{x}+c \\ \frac{x^2}{2}+5x^2- \frac{4}{x}+c \\ \frac{x^2}{2}+5x-\large \frac{4}{x}+c.\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\int x^n dx=\large\frac{x^{n+1}}{n+1}+c\;$
$\large\int\frac{x^3+5x^2-4}{x^2}$$dx$
We can rewrite this as,
$\qquad=\int\frac{x^3}{x^2}dx+\int\frac{5x^2}{x^2}dx-\int\frac{4}{x^2}dx.$
$\qquad=\int xdx+5\int dx-4\int\frac{1}{x^2}dx$
$\qquad=\int xdx+5\int dx-4\int x^{-2}dx.$
$\qquad=\large\frac{x^2}{2}$$+5x-4\big(\large\frac{x^{-1}}{-1}\big)+c.$
$\qquad=\large\frac{x^2}{2}$$+5x+\big(\large\frac{4}{x}\big)+c.$
Hence $\large\int\frac{x^3+5x^2-4}{x^2}$$dx=\large\frac{x^2}{2}$$+5x+\big(\large\frac{4}{x}\big)+c.$
answered Sep 6, 2013 by meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...