Question

Evaluating Integrals

7.Integral of (3c^2x^2-d^4)^2 dx
8. Integral of (square root of 2x + 2x square root of x + 1/ square root of x) dx
9. Integral of t^3 + 2t^2- 3 / cube root of t

Answer 1

Evaluating Integrals

We apply a formula

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

$7 ). \space \space \int (3 c^2 x^2 - d^4 )^2 \space dx$

  (Expand this )

$$=\int ( 9 c^4 x^4 - 6c^2 d^4 x^2 +d^8 ) dx$$

$$=9 c^4 \int x^4 dx - 6c^2 d^4 \int x^2 \space dx +d^8 \int dx + k$$

$$= 9c^4 \frac {x^5}{5} - 6c^2 d^4 \frac {x^3}{3} + d^8 x + k$$

Answer :

$$= 9c^4 \frac {x^5}{5} - 2 c^2 d^4 {x^3} + d^8 \space x + k$$

$8). \space \space \int (\sqrt {2x} + 2x \sqrt x \space + \frac {1} {\sqrt x} ) \space dx$

$$=\sqrt 2 \int \sqrt x \space dx + 2 \int x \sqrt x \space dx + \int \frac {1} {\sqrt x} \space dx + c$$

$$=\sqrt 2 \int x^{\frac {1}{2}} \space dx + 2 \int x \times x^\frac {1}{2} \space dx + \int { x^{(-\frac {1}{2})}} \space dx + c$$

$$=\sqrt 2 \int x^{\frac {1}{2}} \space dx + 2 \int x^\frac {3}{2} \space dx + \int { x^{(-\frac {1}{2})}} \space dx + c$$

Answer:

$$= \sqrt2 \frac {x^{(\frac {1}{2} +1)}}{(\frac {1}{2} +1)} + 2 \frac {x^{(\frac {3}{2} +1)}}{(\frac {3}{2} +1)} + \frac {x^{(-\frac {1}{2} +1)}}{(-\frac {1}{2} +1)} + c$$

9).

$\int ( t^3 + 2 t^2 - 3 t^ {\frac {1}{3}}) \space dt$ =

$$\int t^3 dt + 2 \int t^2 dt - 3 \int t^ {\frac {1}{3}} \space dt + c$$

$$= \frac {t^{(3 +1)}}{(3 +1)} + 2 \frac {t^{({2} +1)}}{({2} +1)} - 3 \frac {t^{(\frac {1}{3} +1)}}{(\frac {1}{3} +1)} + c$$

Answer:

$$= \frac {t^{4}}{4} + 2 \frac {t^{{3}}}{3} - 9 \frac {t^{(\frac {4}{3} )}}{( {4})} + c$$