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Integration by parts Quiz 2

mathematics

 Solve the integral $\int{\left( 9x+\dfrac{9}{x} \right)\left( 9-\dfrac{9}{{{x}^{2}}} \right).  {{e}^{9x+\frac{9}{x}}}}  dx$.

 Substitute $9x+\dfrac{9}{x}=t$ and then use integration by parts.

$\text{Consider the given integral as “I”.} \\  I=\int{\left( 9x+\dfrac{9}{x} \right)\left( 9-\dfrac{9}{{{x}^{2}}} \right){{e}^{9x+\frac{9}{x}}}dx.} \\  \text{Put }9x+\dfrac{9}{x}=t \\ \Rightarrow  \left( 9-\dfrac{9}{{{x}^{2}}} \right)dx=dt \\ I = \int{t{{e}^{t}}  dt} \\ <br>\text{Apply integration by parts } \\  I=t\int{{{e}^{t}} dt  -\int{\left( \dfrac{d}{dt} \left( t \right). \int{{{e}^{t}} dt} \right) dt}} \\ \Rightarrow t{{e}^{t}} - \int{1\times {{e}^{t}} dt} \\ \Rightarrow t{{e}^{t}} - {{e}^{t}} +c \\  \text{Substitute }t = 9x+\dfrac{9}{x} \\ I = \left( 9x+\dfrac{9}{x} \right){{e}^{9x+\frac{9}{x}}} -  {{e}^{9x+\frac{9}{x}}}+c$

Write the value of integral $\int{\left( 14x + 11 \right) \log  \left( 7{{x}^{2}} + 11x + 5 \right)} dx$.

 Put $7{{x}^{2}}+11x+5=t$ and then use by parts. 

Consider the given integral as $“I”$. $ \\  I=\int{\left( 14x+11 \right) \log  \left( 7{{x}^{2}}+11x+5 \right) dx} \\ \text{Put }7{{x}^{2}}+11x+5=t \\ \Rightarrow \left( 14x+11 \right) dx = dt \\ \Rightarrow I = \int{\log   t  dt} \\ $ Take $\log t$ as the first function and 1 as second function and apply by parts.$ \\  \log  t\int{1 dt  -\int{\left( \dfrac{d}{dt}\left( \log  t \right) \int 1 dt \right) dt}} \\ \Rightarrow  I =t \log  t - \int{\dfrac{1}{t} \times t  dt} \\ \Rightarrow I= t \log  t -t +c \\  \text{Put }t = \left( 7{{x}^{2}}+11x+5 \right) \log  \left( 7{{x}^{2}} + 11x+5 \right)-\left( 7{{x}^{2}}+11x+5 \right)+c$ 

 Write the value of integral $\int{{{29}^{m+51}}}. m   dm$.

 $\text{Consider the given integral as “D”.} \\ I=\int{{{29}^{m+51}}mdm} \\  \text{Use integration by parts by taking } \\ m\int{{{29}^{m+51}}dm-\int{\left( \dfrac{d}{dm}\left( m \right)\int{{{29}^{m+51}}dm} \right)dm}} \\ \Rightarrow I=\dfrac{m{{.29}^{m+51}}}{\log 29}-\int{1\times \dfrac{{{29}^{51+m}}}{\log 9}dm} \\  I=\dfrac{m.29}{\log 29}-\dfrac{{{29}^{m+51}}}{{{\left( \log 29 \right)}^{2}}}+c$

Evaluate and write the value of integral $\int{\left( 2{{x}^{3}}+30x \right)  \sin  \left( {{x}^{2}}+15 \right)} dx$.

 Take $2x$ common and then use substitution method and integration by parts.

 $\text{Consider the given integral}  \\  I=\int{\left( 2{{x}^{3}}+30x \right)\sin \left( {{x}^{2}}+15 \right)dx} \\ \Rightarrow \int{2x\left( {{x}^{2}}+15 \right)\sin \left( {{x}^{2}}+15 \right)dx} \\  \text{Put }{{x}^{2}}+15=t \\ \Rightarrow I=\int{{{t}^{\sin t}}dt} \\ \text{Use integration by parts.}  \\  I=t\int{\operatorname{sint dt}-\int{\left( \dfrac{d}{dt}\left( t \right).\int{\sin tdt} \right)dt}} \\ \Rightarrow -t\cos t-\int{1\times -\cos tdt} \\ \Rightarrow -t\cos t+\sin t+c \\ \text{Substitute }t={{x}^{2}}+15 \\ \Rightarrow I=-\left( {{x}^{2}}+15 \right)\cos \left( {{x}^{2}}+15 \right)+\sin \left( {{x}^{2}}+15 \right)+c$

 Simplify the integral $\int{\left( 23p + 5 \right)}  {{e}^{23p+5}} dp$.

 Use integration by parts by taking $23p+5$ as the first function and  ${{e}^{23p+5}}$ as the second function.

 $\text{Let us take the given integral} \\  I=\int{\left( 23p+5 \right){{e}^{23p+5}}dp} \\  \text{Use integration by parts } \\  I=\left( 23p+5 \right).\int{{{e}^{23p+5}}dp-\int{\left( \dfrac{d}{dp}\left( 23p+5 \right).\int{{{e}^{23p+5}}dp} \right)dp}} \\ \Rightarrow \dfrac{\left( 23p+5 \right){{e}^{23p+5}}}{23}-\int{23\times \dfrac{{{e}^{23p+5}}}{23}dp} \\ \Rightarrow \dfrac{\left( 23p+5 \right){{e}^{23p+5}}}{23}-\dfrac{{{e}^{23p+5}}}{23}+c$