Question-50158

Question Number 50158 by cesar.marval.larez@gmail.com last updated on 14/Dec/18

Commented by cesar.marval.larez@gmail.com last updated on 14/Dec/18

i a m p r a c t i c i n g v e r y h a r d

Answered by afachri last updated on 14/Dec/18

(4) \int \frac{2 x^{2} + 8 x + 6}{x - 3} d x = \dots .

let x - 3 = u \Rightarrow d x = d u

x = u + 3 then subtitute to eq;

\int \frac{2 {(u + 3)}^{2} + 8 (u + 3) + 6}{u} d u = \int \frac{2 u^{2} + 20 u + 48}{u}

= \int 2 u + 20 + \frac{48}{u} d u

= u^{2} + 20 u + 48 \ln u + C

= {(x - 3)}^{2} + 20 (x - 3) + 48 \ln (x - 3) + C

= x^{2} + 14 x - 51 + 48 \ln (x - 3) + C

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

4) \int \frac{2 x^{2} + 3 x + 6}{x - 3} d x

t = x - 3

\int \frac{2 {(t + 3)}^{2} + 3 (t + 3) + 6}{t} d t

\int \frac{2 t^{2} + 12 t + 18 + 3 t + 9 + 6}{t} d t

\int \frac{2 t^{2} + 15 t + 33}{t} d t

2 \int t d t + 15 \int d t + 33 \int \frac{d t}{t}

t^{2} + 15 t + 33 l n t + c

{(x - 3)}^{2} + 15 (x - 3) + 33 l n (x - 3) + c

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

5) \int {c o s}^{4} 6 x (- 6 s i n 6 x) d x

t = c o s 6 x d t = - 6 s i n 6 x d x

\int t^{4} d t

\frac{t^{5}}{5} + c

\frac{{(c o s 6 x)}^{5}}{5} + c

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

6) \int {s i n}^{3} x d x

\int (1 - {c o s}^{2} x) s i n x d x

t = c o s x d t = - s i n x d x

\int (1 - t^{2}) \times - d t

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

= \frac{t^{3}}{3} - t + c

= \frac{{(c o s x)}^{3}}{3} - c o s x + c

Commented by cesar.marval.larez@gmail.com last updated on 14/Dec/18

m y f r i e n d i s {s e n}^{3} a x

Commented by afachri last updated on 14/Dec/18

you can solve it over and do like

Mr Chaudri did . and the result

is :

(\frac{\cos^{3} (a x) - 3 \cos (a x)}{3 a}) + C not too far

{isn}^{'} t it ? ?

{don}^{'} t be doubt to try my friend .

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

\int {c o s}^{2} \frac{x}{2} d x

\int \frac{1 + c o s x}{2} d x

\frac{1}{2} \int (1 + c o s x) d x

= \frac{1}{2} (x + s i n x) + c

Commented by cesar.marval.larez@gmail.com last updated on 14/Dec/18

I k n e w t h i s b u t i w a s n o t s e c u r e

t h a n k u s i r h a h a

Answered by afachri last updated on 15/Dec/18

(9) \int 5 x . e^{\frac{4 x}{5}} d x = \dots

Can be solved with partial integral;

\int u d v = u v - \int v d u

let u = 5 x \Rightarrow d u = 5 d x

d v = e^{\frac{4 x}{5}} d x \Rightarrow v = \frac{5}{4} e^{\frac{4 x}{5}}

so u v - \int v d u = (5 x . \frac{5}{4} e^{\frac{4 x}{5}}) - \int \frac{5}{4} e^{\frac{4 x}{5}} (5 d x)

= \frac{25 x}{4} e^{\frac{4 x}{5}} - (\frac{5 \times 5}{4 \times 4} e^{\frac{4 x}{5}} . 5) + C

= \frac{25 e^{\frac{4 x}{5}}}{4} (x - \frac{5}{4}) + C

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18

7) \int 2 x^{3} e^{- \frac{x}{3}} d x

2 x^{3} \int e^{- \frac{x}{3}} d x - \int [\frac{d}{d x} (2 x^{3}) \int e^{- \frac{x}{3}} d x] d x

= 2 x^{3} \times \frac{e^{\frac{- x}{3}}}{\frac{- 1}{3}} - \int 6 x^{2} \times \frac{e^{- \frac{x}{3}}}{\frac{- 1}{3}} d x

= - 6 x^{3} e^{\frac{- x}{3}} + 18 I_{2}

I_{2} = \int x^{2} e^{\frac{- x}{3}} d x

= x^{2} \times \frac{e^{\frac{- x}{3}}}{\frac{- 1}{3}} - \int [\frac{{d x}^{2}}{d x} \int e^{\frac{- x}{3}} d x] d x

= - 3 x^{2} e^{\frac{- x}{3}} - \int 2 x \times \frac{e^{\frac{- x}{3}}}{\frac{- 1}{3}} d x

= - 3 x^{2} e^{\frac{- x}{3}} + 6 \int {x e}^{\frac{- x}{3}} d x

= - 3 x^{2} e^{\frac{- x}{3}} + 6 I_{3}

I_{3} = \int {x e}^{\frac{- x}{3}} d x

= x \times \frac{e^{- \frac{x}{3}}}{\frac{- 1}{3}} - \int [\frac{d x}{d x} \int e^{\frac{- x}{3}} d x] d x

= \frac{- 3 {x e}^{\frac{- x}{3}}}{1} - \int \frac{e^{\frac{- x}{3}}}{\frac{- 1}{3}} d x

= \frac{- 3 {x e}^{\frac{- x}{3}}}{1} + 3 \times \frac{e^{\frac{- x}{3}}}{\frac{- 1}{3}} + C

= - 3 {x e}^{\frac{- x}{3}} - 9 e^{- \frac{x}{3}} + C

p l s d o r e s t b y p u t t i n g I_{2} a n d I_{3}