Evaluate the following integrals. ∫exe2x+2ex+1dx

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Evaluate the following integrals.  ∫exe2x+2ex+1dx

kalupunangh · Accepted Answer

Step 1  We have to evaluate the integral:  ∫exe2x+2ex+1dx  Rewriting the integral,  ∫exe2x+2ex+1dx=∫ex(ex)+2ex+1dx  This will be solved by substitution method,  so assuming t=ex  differentiating,  dtdx=dexdx  dtdx=ex  dt=exdx  Step 2  Substituting above values in the integral,  ∫ex(ex)2+2ex+1dx=∫dtt2+2t+1  =∫dt(t)2+2(t)(1)+(1)2  =∫dt(t+1)2  =−1t+1+C   (since ∫1(x+a)2=−1x+a+C)  Where, C is an arbitrary constant.  Hence, value of integral is −1ex+1+C.

Barbara Meeker · Accepted Answer

It is required to calculate:  ∫exe2x+2ex+1dx  Substitution u=ex⇒dudx=ex⇒dx=e−xdu, we use:  e2x=u2  =∫1u2+2u+1du  Let us factorize:  =∫1(u+1)2du  Substitution v=u+1⇒dvdu=1⇒du=dv:  =∫1v2dv  Integral of a power function:  ∫vndv=vn+1n+1 at n=-2:  =−1v  Reverse replacement v=u+1:  =−1u+1  Reverse replacement u=ex:  =−1ex+1  Problem solved:  ∫exe2x+2ex+1dx  =−1ex+1+C

nick1337 · Accepted Answer

∫exe2x+2ex+1dx Let us put the expression exp (x) under the sign of the differential, ie: ex∗dx=d(ex),t=ex Then the original integral can be written as follows: ∫1t2+2∗t+1∗dt ∫1x2+2∗x+1∗dx Let&#039;s 1(x+1)2=1(x+1)2 use the method of decomposition into simplest elements. Let us expand the function into the simplest terms: 1(x+1)2=Ax+1+B(x+1)2=A(x+1)+B(x+1)2 Equate the numerators and take into account that the coefficients at the same powers of x , standing on the left and on the right, must coincide: 1=A(x+1)+B x: A=0 1: A+B=1 Solving it, we find: A=0; B=1; 1(x+1)2=0x+1+1(x+1)2 We calculate the table integral: ∫1(x+1)2∗dx=−1x+1 NSK Answer: −1x+1+C To write down the final answer, it remains to substitute exp (x) instead of t. −1ex+1+C

Evaluate the following integrals. \int \frac{e^{x}}{e^{2x}+2e^{x}+1}dx

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