Question

Let I=∫(2x+3)23​7+4x+5x2​.dx =(2x+3)23​5x2+4x+7​.dx Put 2x+3=t ∴2dx=dt ∴dx=2dt​ Also, x=2t−3​ ∴I=∫t23​5(2t−2​)2+4(2t−3​)+7​.2dt​ =21​∫t23​5(4t2−6t+9​)+2(t−3)+7​dt =21​∫4t23​5t2−30t+45+8t−24+28​dt =81​∫t23​5t2−22t+49​dt =81​∫⎝⎛​5t21​−22t−21​+49t−23​⎠⎞​dt 85​∫t21​dt−822​∫t−21​dt+849​∫t−23​dt =85​.⎝⎛​t−23​⎠⎞​t−23​​−411​.⎝⎛​t−21​⎠⎞​t−21​​+849​.⎝⎛​t−23​⎠⎞​t−21​​+c 125​(x+3)23​−211​2x+3​−449​.2x+3​1​+c​

Answer 1

Step-by-step explanation:

Let I=∫

(2x+3)

2

3

7+4x+5x

2

.dx

=

(2x+3)

2

3

5x

2

+4x+7

.dx

Put 2x+3=t

∴2dx=dt

∴dx=

2

dt

Also, x=

2

t−3

∴I=∫

t

2

3

5(

2

t−2

)

2

+4(

2

t−3

)+7

.

2

dt

=

2

1

∫

t

2

3

5(

4

t

2

−6t+9

)+2(t−3)+7

dt

=

2

1

∫

4t

2

3

5t

2

−30t+45+8t−24+28

dt

=

8

1

∫

t

2

3

5t

2

−22t+49

dt

=

8

1

∫

⎝

⎛

5t

2

1

−22t

−

2

1

+49t

−

2

3

⎠

⎞

dt

8

5

∫t

2

1

dt−

8

22

∫t

−

2

1

dt+

8

49

∫t

−

2

3

dt

=

8

5

.

⎝

⎛

t

−

2

3

⎠

⎞

t

−

2

3

−

4

11

.

⎝

⎛

t

−

2

1

⎠

⎞

t

−

2

1

+

8

49

.

⎝

⎛

t

−

2

3

⎠

⎞

t

−

2

1