Integrate {1+x}{x³} dx

Correct Answer is : −12x2−1x+k- {1}{2x²} - 1/x + k−2x21−x1+k

Correct Answer is : −12x2−1x+k- {1}{2x²} - 1/x + k−2x21−x1+k

Test Index

JAMB Mathematics 2013 Paper

Question : 46

Total: 47

Integrate

\frac{1+x}{x^{3}} dx

Solution:

Concept:

Rewrite the integrand as a sum of powers of

x

and integrate each term using the power rule.

Explanation:

Start by dividing each term in the numerator by

x^{3}

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

Now integrate the two terms separately:

\int (x^{-3} + x^{-2})\,dx = \int x^{-3}\,dx + \int x^{-2}\,dx

Apply the power rule

\int x^{n}\,dx = \frac{x^{n+1}}{n+1} + k

for

n \neq -1

\int x^{-3}\,dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^{2}}

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

Combine the results and add the constant of integration

k

\int \frac{1+x}{x^{3}}\,dx = -\frac{1}{2x^{2}} - \frac{1}{x} + k

Answer:

-\frac{1}{2x^{2}} - \frac{1}{x} + k

(Option B).

Go to Question:

Prev Question

Next Question