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Integration by Parts

Total questions: 10

Worksheet time: 30mins

Name
Class
Date
1.

Evaluate the indefinite integral using integration by parts. 
 ∫3x e2x dx \int3x\ e^{2x}\ dx\   

a)

 −xe2x2+e2x4+C-\frac{xe^{2x}}{2}+\frac{e^{2x}}{4}+C  

b)

 3xe2x2−3e2x4+C\frac{3xe^{2x}}{2}-\frac{3e^{2x}}{4}+C  

c)

 xe−2x+(1−x2)2+Cxe^{-2x}+\frac{\left(1-x^2\right)^{ }}{2}+C  

d)

 −xe2x2+ln⁡e2x4+C-\frac{xe^{2x}}{2}+\frac{\ln e^{2x}}{4}+C  

2.

Evaluate the indefinite integral using integration by parts. 
 ∫t2ln⁡t dt \int t^2\ln t\ dt\   

a)

 2t2ln⁡2t−t24+C\frac{2t^2\ln2t-t^2}{4}+C  

b)

 t3 ln⁡33−t39+C\frac{t^3\ \ln3}{3}-\frac{t^3}{9}+C  

c)

 et2t+2+C\frac{e^t}{2t+2}+C  

d)

 −2t−14e2t+C\frac{-2t-1}{4e^{2t}}+C  

3.

 ∫xsin⁡(8x)dx\int x\sin\left(8x\right)dx  

a)

 =−x8cos⁡(8x)+164sin⁡(8x)+C=-\frac{x}{8}\cos\left(8x\right)+\frac{1}{64}\sin\left(8x\right)+C  

b)

 =−x8cos⁡(8x)−164sin⁡(8x)+C=-\frac{x}{8}\cos\left(8x\right)-\frac{1}{64}\sin\left(8x\right)+C  

c)

 =−x8cos⁡(8x)+18sin⁡(8x)+C=-\frac{x}{8}\cos\left(8x\right)+\frac{1}{8}\sin\left(8x\right)+C  

d)

 =x8cos⁡(8x)+164sin⁡(8x)+C=\frac{x}{8}\cos\left(8x\right)+\frac{1}{64}\sin\left(8x\right)+C  

4.

 ∫xe−xdx\int xe^{-x}dx  

a)

 =−e−x(x−1)+C=-e^{-x}\left(x-1\right)+C  

b)

 =e−x(x+1)+C=e^{-x}\left(x+1\right)+C  

c)

 =e−x(x−1)+C=e^{-x}\left(x-1\right)+C  

d)

 =−e−x(x+1)+C=-e^{-x}\left(x+1\right)+C  

5.

Evaluate the indefinite integral using integration by parts. 
 ∫log⁡5t dt ;   \int\log_5t\ dt\ ;\ \ \   

a)

 2t2ln⁡3t−t24+C\frac{2t^2\ln3t-t^2}{4}+C  

b)

 −t3tln⁡3−13t⋅(ln⁡3)2+C-\frac{t}{3^t\ln3}-\frac{1}{3^t\cdot\left(\ln3\right)^2}+C  

c)

 tln⁡(t2+9)−2t+6tan⁡−1(t3)+Ct\ln\left(t^2+9\right)-2t+6\tan^{-1}\left(\frac{t}{3}\right)+C  

d)

 tlog⁡5t−tln⁡5+Ct\log_5t-\frac{t}{\ln5}+C  

6.

 ∫x2sin⁡x dx\int_{ }x^2\sin x\ dx  by using integration by parts.

a)

 xsin⁡x− ex+Cx\sin x-\ e^x+C  

b)

 (−excos⁡x−exsin⁡x+2) +C\left(\frac{-e^x\cos x-e^x\sin x+}{2}\right)\ +C  

c)

 xex+Cxe^x+C  

d)

 xex−ex+Cxe^x-e^x+C  

7.

Evaluate the indefinite integral using integration by parts
 ∫xe4xdx ;  \int xe^{4x}dx\ ;\ \   

a)

 −x4xln⁡4−14x⋅(ln⁡4)2+C-\frac{x}{4^x\ln4}-\frac{1}{4^x\cdot\left(\ln4\right)^2}+C  

b)

 x5ln⁡x5−x525+C\frac{x^5\ln x}{5}-\frac{x^5}{25}+C  

c)

 xln⁡(x+4)−x+4ln⁡(x+4)+Cx\ln\left(x+4\right)-x+4\ln\left(x+4\right)+C  

d)

 xe4x4−e4x16+C\frac{xe^{4x}}{4}-\frac{e^{4x}}{16}+C  

8.

Evaluate the indefinite integral using integration by parts
 ∫e4xcos⁡(2x)dx \int e^{4x}\cos\left(2x\right)dx\   

a)

 e4xsin⁡(2x)10+e4xcos⁡(2x)5+C\frac{e^{4x}\sin\left(2x\right)}{10}+\frac{e^{4x}\cos\left(2x\right)}{5}+C  

b)

 −e4xsin⁡(2x)10+e4xcos⁡(2x)5+C-\frac{e^{4x}\sin\left(2x\right)}{10}+\frac{e^{4x}\cos\left(2x\right)}{5}+C  

c)

 e4xcos⁡(2x)10+e4xsin⁡(2x)5+C\frac{e^{4x}\cos\left(2x\right)}{10}+\frac{e^{4x}\sin\left(2x\right)}{5}+C  

d)

 e4xsin⁡2(2x)10+C\frac{e^{4x}\sin^2\left(2x\right)}{10}+C  

9.

Evaluate the indefinite integral using integration by parts
 ∫e2xdx\int e^{\sqrt{2x}}dx 

a)

 2e2xx+e2x+C\sqrt{2}e^{\sqrt{2}\sqrt{x}}\sqrt{x}+e^{\sqrt{2}\sqrt{x}}+C 

b)

 2e2xx+C2e^{\sqrt{2}\sqrt{x}}\sqrt{x}+C 

c)

 e2xx−e2x+Ce^{\sqrt{2}\sqrt{x}}x-e^{\sqrt{2}\sqrt{x}}+C 

d)

 2e2xx−e2x+C\sqrt{2}e^{\sqrt{2}\sqrt{x}}\sqrt{x}-e^{\sqrt{2}\sqrt{x}}+C 

10.

Evaluate the indefinite integral using integration by parts
 ∫sin⁡(x)dx\int\sin\left(\sqrt{x}\right)dx 

a)

 −2xsin⁡2(x)+C-\sqrt{2}\sqrt{x}\sin^2\left(\sqrt{x}\right)+C 

b)

 −2xcos⁡(x)+2sin⁡(x)+C-2\sqrt{x}\cos\left(\sqrt{x}\right)+2\sin\left(\sqrt{x}\right)+C 

c)

 2xcos⁡(x)+sin⁡x+C\sqrt{2}\sqrt{x}\cos\left(\sqrt{x}\right)+\sin\sqrt{x}+C 

d)

 2(−xcos⁡(x)−sin⁡(x))+C\sqrt{2}\left(-\sqrt{x}\cos\left(\sqrt{x}\right)-\sin\left(\sqrt{x}\right)\right)+C