Indefinite Integration : Simple Problems

In this blog, few simple problems on indefinite integration have been solved. All you need to know is the basic formulas for integration (the opposite of differentiation) and the algebra of integration of functions, i.e. how to integrate the addition and subtraction of 2 functions and the integration of simple composite functions of the form $f(ax+b)$.

$\int 3sec^2(x) - \frac {4}{x} + \frac {1}{x \sqrt {x}} - 7 \ dx$

$= \int 3sec^2(x)dx - \int \frac {4}{x}dx + \int \frac {1}{x \sqrt {x}}dx - \int 7 dx$

$= 3 \int sec^2(x)dx - 4 \int \frac {1}{x} dx + \int x^{-3/2} dx - 7 \int 1 \ dx$

$= 3 tan(x) - 4 ln(x) + \frac {x^{- \frac {3}{2} +1 }}{- \frac {3}{2} + 1} - 7x + C$

${= 3 tan(x) - 4 ln(x) - \frac {2}{\sqrt {x}} - 7x + C}$

$\int \frac {1}{(7x-5)^3} - \frac {1}{\sqrt {5x-4}} \ dx$

$= \int (7x-5)^{-3}dx - \int (5x-4)^{-1/2} dx$

$= \frac {1}{7} \frac {(7x-5)^{-3+1}}{-3+1} - \frac {1}{5} \frac {(5x-4)^{- \frac {1}{2} + 1}}{- \frac 1 2 + 1} + C$

${= \frac {-1}{14} \frac {1}{(7x-5)^2} - \frac {2}{5} \sqrt {5x-4} + C}$

$\int \frac {1}{\sqrt {5x+3} - \sqrt {5x+1}} dx$

$= \int \frac {\sqrt {5x+3} + \sqrt {5x+1}}{5x+3 - (5x+1)} dx$

$= \int \frac {\sqrt {5x+3} + \sqrt {5x+1}}{2}$

$= \frac {1}{2} \int \sqrt {5x+3}dx + \frac {1}{2} \int \sqrt {5x+1} dx$

$= \frac {1}{2} \cdot \frac {1}{5} \cdot (5x+3)^{3/2} \cdot \frac {2}{3} + \frac {1}{2} \cdot \frac {1}{5} \cdot (5x+1)^{3/2} \cdot \frac {2}{3} + C$

${= \frac {(5x+3)^{3/2}}{15} + \frac {(5x+1)^{3/2}}{15} + C}$

$\int \frac {3x^2-2x+5}{x^{3/2}} dx$

$= \int \frac {3x^2}{x^{3/2}} dx - \frac {2x}{x^{3/2}} dx + \int \frac {5}{x^{3/2}} dx$

$= 3 \int x^{2 - \frac 3 2} dx - 2 \int x^{1 - \frac 3 2} dx + 5 \int x^{-3/2} dx$

$= 3 \int {x^{1/2}} dx - 2 \int x^{-1/2} dx + 5 \int x^{-3/2} dx$

${= 2 x^{3/2} - 4 \sqrt {x} - \frac {10}{\sqrt {x}} + C}$

1. Hey thanks Rohan! The crucial part in this problem is to rationalize the denominator of the integrand. So, you should multiply and divide by $\sqrt {5x+3} + \sqrt {5x+1}$, which is done in the very step. The radical sign goes when you square and $5x$ terms go. Hence $2$ in the denominator.